Answer:
19.872 degrees
Explanation:
Mathematically;
Using Snell’s law
n1 sin A = n2 sinB
Where ;
n1 = refractive index in air = 1
n2 is refractive index in water = 1.33
A = ?
B = 45
Substituting the values in the equation;
1 sin A = 1.33 sin45
Sin A = 1.33 sin 45
A = arc sin (1.33 sin 45)
A = 70.12
Thus, the actual direction of the Sun with respect to the horizon = 90-A = 19.872 degrees
What is the power of a child that has
done work of 50J in 10 seconds.
(a)50W (b)20W (c)30W (d)5W
_____________________________
Solution,
Work=50 Joule
Time=10 seconds
Power=?
Now,
Power=Work/time
= 50/10
= 5 Watt.
So the right answer is 5 W
Hope it helps..
Good luck on your assignment
__________________________
A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static friction, to prevent the skater from slipping sideways, but Uk = 0. Suddenly, a wind from the northeast exerts a force of 3.70 N on the skater.a) Use work and energy to find the skater's speed after gliding 100 m in this wind.b) What is the minimum value of Ug that allows her to continue moving straight north?
Answer:
a. 2.668 m/s
b. 0.00494
Explanation:
The computation is shown below:
a. As we know that
[tex]W = F\times d[/tex]
[tex]KE = 0.5\times m\times v^2[/tex]
As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.
F = 3.70 cos 45 = 2.62 N
[tex]W = F \times d = 2.62 N \times 100 m[/tex]
[tex]W = 261.6 N\times m[/tex]
We know that
KE1 = Initial kinetic energy
KE2 = kinetic energy following 100 m
The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.
So, the equation is
KE2 = KE1 - W
[tex]0.5 m\times v2^2 = 0.5 m\ v1^2 - W[/tex]
Now solve for v2
[tex]v2 = \sqrt{v1^2 - {\frac{2W}{M}}}[/tex]
[tex]= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}[/tex]
= 2.668 m/s
b. Now the minimum value of Ug is
As we know that
Ff = force of friction
Us = coefficient of static friction
N = Normal force = weight of skater
So,
[tex]Ff = Us\times N[/tex]
Now solve for Us
[tex]= \frac{Ff}{N}[/tex]
[tex]= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}[/tex]
= 0.00494
I really need help with this question someone plz help !
Answer:D
Explanation:
Given
Same force is applied to each ball such that all have different masses
and Force is given by the product of mass and acceleration
[tex]F=m\times a[/tex]
[tex]a=\frac{F}{m}[/tex]
So acceleration of ball A
[tex]a_A=\frac{F}{0.5}=2F[/tex]
acceleration of ball B
[tex]a_B=\frac{F}{0.75}=\frac{4F}{3}=1.33F[/tex]
acceleration of ball C
[tex]a_C=\frac{F}{1}=F[/tex]
acceleration of ball D
[tex]a_D=\frac{F}{7.3}=\frac{F}{7.3}[/tex]
It is clear that acceleration of ball D is least.
The NASA spacecraft Deep Space I was shut down on December 18, 2001, following a three-year journey to the asteroid Braille and the comet Borrelly. This spacecraft used a solar-powered ion engine to produce 0.064 ounces of thrust (force) by stripping electrons from neon atoms and accelerating the resulting ions to 70,000 mi/h. The thrust was only as much as the weight of a couple sheets of paper, but the engine operated continuously for 16,000 hours. As a result, the speed of the spacecraft increased by 7900 mi/h. What was the mass of Deep Space I?
Answer:
The mass will be "8.86 lb".
Explanation:
The given values are:
Force
= 70,000 mi/h
Speed
= 7900 mi/h
On applying the Law of momentum, we get
⇒ [tex]V_{1}m_{1}=V_{2}m_{2}[/tex]
On putting the estimated values, we get
⇒ [tex]70000 = 7900\times mass \ of \ deepspace \ 1[/tex]
⇒ [tex]mass \ of \ deepspace \ 1 = \frac{70000}{7900}[/tex]
⇒ [tex]=8.86 \ lb[/tex]
A projectile is defined as
Answer:
By definition, a projectile has a single force that acts upon it - the force of gravity.
Explanation:
A projectile is any object that once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.
// have a great day //
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.
Answer:
Letter A. [tex]y=4.55 m[/tex]
Explanation:
Let's use the wave equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction (x=1.21 m)Therefore the displacement y will be:
[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]
[tex]y=4.55 m[/tex]
The answer is letter A.
I hope it helps you!
Answer:
Explanation:
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
Amplitude (A) of the simple harmonic wave = 6.44 m
wave number (k) of the given wave = 2.34 m-1
Angular frequency (ω) of the given wave = 2.88 rad/s
Displacement x = 1.21 m and time t = 0.71 s
Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by
y = Asin(kx - ωt)
= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]
Y=6.44sin(0.7866 rad)
0.7866rad*(180 degrees/pi rad) =45.1
Y=6.44sin(45.1)
Y=4.55m
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
The mass of the block is 1250g.
Explanation:
Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :
[tex]ρ = \frac{mass}{volume} [/tex]
Let density = 250,
Let volume = 5,
[tex]250 = \frac{m}{5} [/tex]
[tex]m = 250 \times 5[/tex]
[tex]m = 1250g[/tex]
A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her
Answer:
The average upward force exerted by the water is 988.2 N
Explanation:
Given;
mass of the diver, m = 60 kg
height of the board above the water, h = 10 m
time when her feet touched the water, t = 2.10 s
The final velocity of the diver, when she is under the influence of acceleration of free fall.
V² = U² + 2gh
where;
V is the final velocity
U is the initial velocity = 0
g is acceleration due gravity
h is the height of fall
V² = U² + 2gh
V² = 0 + 2 x 9.8 x 10
V² = 196
V = √196
V = 14 m/s
Acceleration of the diver during 2.10 s before her feet touched the water.
14 m/s is her initial velocity at this sage,
her final velocity at this stage is zero (0)
V = U + at
0 = 14 + 2.1(a)
2.1a = -14
a = -14 / 2.1
a = -6.67 m/s²
The average upward force exerted by the water;
[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]
Therefore, the average upward force exerted by the water is 988.2 N
Proposed Exercise - Mass Center of a Composite Body Determine the coordinates (x, y) of the center of mass of the body illustrated in the picture below
Answer:
x = 3.76 cm
y = 3.76 cm
Explanation:
This composite shape can be modeled as a square (7.2 cm × 7.2 cm) minus a quarter circle in the lower left corner (3.6 cm radius) and a right triangle in the upper right corner (3.6 cm × 3.6 cm).
The centroid of a square (or any rectangle) is at x = b/2 and y = h/2.
The centroid of a quarter circle is at x = y = 4r/(3π).
The centroid of a right triangle is at x = b/3 and y = h/3.
Build a table listing each shape, the coordinates of its centroid (x and y), and its area (A). Use negative areas for the shapes that are being subtracted.
Next, multiply each coordinate by the area (Ax and Ay), sum the results (∑Ax and ∑Ay), then divide by the total area (∑Ax / ∑A and ∑Ay / ∑A). The result will be the x and y coordinates of the center of mass.
See attached image.
A conducting sphere contains positive charge distributed uniformly over its surface. Which statements about the potential due to this sphere are true? All potentials are measured relative to infinity
a. The potential at the center of the sphere is zero.
b.The potential is lowest, but not zero, at the center of the sphere.
c. The potential at the center of the sphere is the same as the potential at the surface.
d. The potential at the center is the same as the potential at infinity.
e. The potential at the surface is higher than the potential at the center.
Answer:
a. FALSE
b. FALSE
c. TRUTH
d. FALSE
e. FALSE
Explanation:
To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:
[tex]V=\frac{Q}{4\pi \epsilon_o R}[/tex] inside the sphere
[tex]V'=\frac{Q}{4\pi \epsilon_o r}[/tex] for r > R (outside the sphere)
R: radius of the sphere
ε0: dielectric permittivity of vacuum
Q: charge of the sphere
As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.
Hence, you can conclude:
a. The potential at the center of the sphere is zero. FALSE
b.The potential is lowest, but not zero, at the center of the sphere. FALSE
c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH
d. The potential at the center is the same as the potential at infinity. FALSE
e. The potential at the surface is higher than the potential at the center. FALSE
A low C (f = 65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
Tension of the wire(T) = 169 N
Explanation:
Given:
f = 65Hz
Length of the piano wire (L) = 2 m
Mass density = 5.0 g/m² = 0.005 kg/m²
Find:
Tension of the wire(T)
Computation:
f = v / λ
65 = v / 2L
65 = v /(2)(2)
v = 260 m/s
T = v² (m/l)
T = (260)²(0.005/2)
T = 169 N
Tension of the wire(T) = 169 N
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest uniformly in a distance of 240 m. What are the magnitude and direction of the net force applied to the car to bring it to rest?
Answer:
The value of F= - 830 N
Since the force is negative, it implies direction of the force applied was due south.
Explanation:
Given data:
Mass = 1000-kg
Distance, d = 240 m
Initial velocity, v1 = 20.0 m/s
Final velocity, v2 = 0 (since the car came to rest after brake was applied)
v2²= v1² + 2ad (using one of the equation of motion)
0= 20² + (2 x a x 240)
0= 400 + 480 a
a = - 400/480
a = - 0.83 m/s²
Then, imputing the value of a into
F = ma
F = 1000 kg x ( - 0.83 m/s²)
F= - 830 N
The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.
Blocks of mass 10, 30, and 90 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 32 N is applied to the left-most block. 1) What is the magnitude of the force that the left block exerts on the middle one
Answer:
32N
Explanation:
The Left force exerts an opposite horizontal force of 32N on the middle object
A worker pushes on a crate that experiences a net force of 45.0 N. If it accelerates at 0.500 m/s2 what is the weight?
Answer:
882 N
Explanation:
F = ma
45.0 N = m (0.500 m/s²)
m = 90.0 kg
mg = 882 N
the part of the brain stem called the has been shown to related to arousal in lab animals. when this part is stimulated the animal is awake when it is severed cut the animal goes into coma
Answer:
Its called PSY
Explanation: I so do not know why they named it this way but, hope i helped.
small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? It is impossible to tell since the velocities are not given. The truck experiences the greater average force. It is impossible to tell since the masses are not given. The small car and the truck experience the same average force. The small car experiences the greater average force.
Answer:
The correct option is D: "The small car and the truck experience the same average force."
Explanation:
The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.
Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane
Answer:
5.1 rad/s
Explanation:
Mechanical energy of the system is conserved since no external work is done on the sphere.
[tex]mgh = mv^2/2 + I\omega^2/2[/tex]
Substituting v = ωr and I = 2 m r^2/5, we get,
=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]
=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]
=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]
=> [tex]gh = 7\omega^2 r^2/10[/tex]
=> [tex]\omega r = (10gh/7)^{1/2}[/tex]
=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s
How do you convert 1.3*10^6cal into joules
Answer:
5.4×10⁶J
Explanation:
1 cal = 4.184 J
1.3×10⁶ cal × (4.184 J/cal) = 5.4×10⁶J
A 9.0-V battery (with nonzero resistance) and switch are connected in series across the primary coil of a transformer. The secondary coil is connected to a light bulb that operates on 120 V. Determine the ratio of the secondary to primary turns needed for the bells transformer. Determine the ratio of the secondary to primary turns needed for the bells transformer. Ns/Np=?
Answer:
N₂ / N₁ = 13.3
Explanation:
A transformer is a system that induces a voltage in the secondary due to the variation of voltage in the primary, the ratio of voltages is determined by the expression
ΔV₂ = N₂ /N₁ ΔV₁
where ΔV₂ and ΔV₁ are the voltage in the secondary and primary respectively and N is the number of windings on each side.
In this case, they indicate that the primary voltage is 9.0 V and the secondary voltage is 120 V
therefore we calculate the winding ratio
ΔV₂ /ΔV₁ = N₂ / N₁
N₂ / N₁ = 120/9
N₂ / N₁ = 13.3
s good clarify that in transformers the voltage must be alternating (AC)
Space-faring astronauts cannot use standard weight scales (since they are constantly in free fall) so instead they determine their mass by measuring the period of oscillation when sitting in a chair connected to a spring. Suppose a chair is connected to a spring with a spring constant of 600 N/m. If the empty chair oscillates with a period of 0.9s, what is the mass of an astronaut who oscillates with a period of 2.0 s while sitting in the chair
Answer:
ma = 48.48kg
Explanation:
To find the mass of the astronaut, you first calculate the mass of the chair by using the information about the period of oscillation of the empty chair and the spring constant. You use the following formula:
[tex]T=2\pi\sqrt{\frac{m_c}{k}}[/tex] (1)
mc: mass of the chair
k: spring constant = 600N/m
T: period of oscillation of the chair = 0.9s
You solve the equation (1) for mc, and then you replace the values of the other parameters:
[tex]m_c=\frac{T^2k}{4\pi^2}=\frac{(0.9s)^2(600N/m)}{4\pi^2}=12.31kg[/tex] (2)
Next, you calculate the mass of the chair and astronaut by using the information about the period of the chair when the astronaut is sitting on the chair:
T': period of chair when the astronaut is sitting = 2.0s
M: mass of the astronaut plus mass of the chair = ?
[tex]T'=2\pi\sqrt{\frac{M}{k}}\\\\M=\frac{T'^2k}{4\pi^2}=\frac{(2.0s)^2(600N/m)}{4\pi^2}\\\\M=60.79kg[/tex] (3)
Finally, the mass of the astronaut is the difference between M and mc (results from (2) and (3)) :
[tex]m_a=M-m_c=60.79kg-12.31kg=48.48kg[/tex]
The mass of the astronaut is 48.48 kg
A quartz sphere is 14.0 cm in diameter. What will be its change in volume if its temperature is increased by 305°F? The coefficient of volume expansion of quartz is 1.50×10^6/°C. Answer in cm^3 .
Answer:
0.365 cm³
Explanation:
The change in volume is found by multiplying the coefficient of expansion by the volume and the temperature change. The temperature change is in °F, but the expansion coefficient is per °C, so we need to convert the temperature scale in the computation.
ΔV = V·Ce·ΔT
= (π/6·d³)(1.5×10⁻⁶/°C)((5 °C)/(9 °F))(305 °F)
= (1436.76 cm³)(1.5×10⁻⁶/°C)(169.44 °C)
= 0.365 cm³ . . . . increase in volume
Write the first equation of motion. Under what condition(s) is this equation valid?
Explanation:
The first equation of motion in kinematics is given by :
[tex]v=u+at[/tex] .....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
someone please help me out thanks
Answer:
The answer is D)Theory
Explanation:
This is due because a theory is a scientific term and is a testable model that scientists seek to explain a phenomenon. You can also find out the answer by the process of elimination it can't be data because that would be something they already know and something they use to prove not explain. It can't be law because it isn't testable but can be used to explain. So that leaves you with two answers hypothesis and theory which are very similar but it isn't hypothesis because it isn't used to explain it to help the scientists come up with a theory and accumulate what might happen.
A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object
Answer:
v = 25.45 m/s
Explanation:
In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:
[tex]h_{max}=\frac{v_o^2}{g}[/tex] (1)
vo: initial speed of the object = 18 m/s
g: gravitational acceleration = 9.8 m/s²
Furthermore you use the following formula for the final speed of the object:
[tex]v^2=v_o^2-2gh[/tex] (2)
h: height
You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:
[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]
Then, you solve the previous result for vo:
[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]
The initial speed of the object was 25.45 m/s
. A ball weighs 120g on the earth surface,
i) What is its mass on the surface of the moon? 1mk
Answer:
WEIGHT ON MOON IS 0.2004N
Explanation:
mass of the body=120g=[tex]\frac{120}{1000}[/tex]kg=0.12kg (we will convert g into kg)
gravity on moon=1.67m/s²( to find the mass of anybody on another we should know its gravity)
as we know that (from the formula of weight)
weight=mass×gravity
w=mg
w=0.12kg²×1.67m/s²
w=0.2004N
A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are Group of answer choices down and backwards. down, backwards, and outwards. down and inwards. down and outwards. straight down.
Answer:
down, backwards, and outwards.
Explanation:
For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.
For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.
For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.
The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.
A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west at 5.0 m/s . A wise elder duck finally realizes that the solution is to fly at an angle to the wind.If the ducks can fly at 7.0 m/s relative to the air, what direction should they head in order to move directly south?
The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:
ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)
or equivalently,
[tex]\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}[/tex]
(see the attached graphic)
We have
ducks (relative to wind) = 7.0 m/s in some direction θ relative to the positive horizontal direction, or[tex]\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]
wind (relative to Earth) = 5.0 m/s due East, or[tex]\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)[/tex]
ducks (relative to earth) = some speed v due South, or[tex]\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]
Then by setting components equal, we have
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0[/tex]
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]
We only care about the direction for this question, which we get from the first equation:
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}[/tex]
[tex]\cos\theta=-\dfrac57[/tex]
[tex]\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)[/tex]
or approximately 136º or 224º.
Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want
[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]
which means θ must be between 180º and 360º (since angles in this range have negative sine).
So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.
A small car and an SUV are at a stoplight. The car has a mass equal to half that of the SUV, and the SUV's engine can produce a maximum force equal to twice that of the car. When the light turns green, both drivers floor it at the same time. Which vehicle pulls ahead of the other vehicle after a few seconds?
Complete Question
A small car and an SUV are at a stoplight. The car has amass equal to half that of the SUV, and the SUV's engine can produce a maximum force equal to twice that of the car. When the light turns green, both drivers floor it at the same time. Which vehicle pulls ahead of the other vehicle after a few seconds?
a) It is a tie.
b) The SUV
c) The car
Answer:
The correct option is a
Explanation:
From the question we are told that
The mass of the car is [tex]m_c[/tex]
The force of the car is F
The mass of the SUV is [tex]m_s = 2 m_c[/tex]
The force of the SUV is [tex]F_s = 2 F[/tex]
Generally force of the car is mathematically represented as
[tex]F= m_ca_c[/tex]
[tex]a_c[/tex] is acceleration of the car
Generally force of the car is mathematically represented as
[tex]F_s = m_s * a_s[/tex]
[tex]a_s[/tex] is acceleration of the SUV
=> [tex]2 F = 2 m_c a_s[/tex]
[tex]F = m_c a_s[/tex]
=> [tex]m_c a_s = m_ca_c[/tex]
So [tex]a_s = a_c[/tex]
This means that the acceleration of both the car and the SUV are the same
1. Which of the following is NOT a vector quantity? (a) Displacement. (b) Energy. (c) Force. (d) Momentum. (e) Velocity.
Answer:
B. energy
Explanation:
A vector has direction.
Energy does not have a direction.