One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. The force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00 x 103 m/s, given the collision lasts 6.00 x 10-8 s is Fill input: x 106 N.
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ [tex]F=m(\frac{\Delta v}{\Delta t} )[/tex]
On substituting the given values, we get
⇒ [tex]=0.100\times 10^{-6}(\frac{4\times 10^3}{6\times 10^{-8}} )[/tex]
⇒ [tex]=6666.67 \ N[/tex]
What is the speed of a ball that is attached to a string and swings in a horizontal circle of radius 2.0 m with the central acceleration of 15 m/s^2?
Answer:
5.48 m/s.
Explanation:
Use the formula a=v^2/r.
When does a rubber band, which has been shot at a wall, have the most potential energy?
A When it is lying on the floor.
B When it is flying across the room.
C When it hits the wall.
D When it is stretched ready to shoot at the wall.
Answer:
C because that requires lots of energy
PLEASE ANSWER WITH ACTUAL ANSWER AND I WILL MARK BRAINLIEST (IF YOU GIVE ME A SCAMMY ANSWER I WILL REPORT YOU!!!)
A student wants to determine the local value of the gravitational field strength, g , in their classroom. Which of the following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured?
Select TWO answers.
A: Run a lab cart down an inclined plane; measure the length of the ramp and the time it takes the cart to reach the bottom.
B: Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.
C: Accelerate a lab cart horizontally; measure the mass of the cart and its acceleration.
D: Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.
Answer:
Most likely (B)
Explanation:
B in the passage is the most representative out of all your choices and it has evidence from the passage
Hope dis helps Jit!
Sorry i forgot to type C
B and C both measure mass while the others are calculations and are bias
The following experimental set-ups would allow a student to calculate the magnitude of the gravitational field strength using only the quantities measured:
Hang a known mass from a spring scale; measure the spring scale reading when the mass is at rest.Drop a heavy metal ball; measure the drop height and the time it takes the ball to hit the ground.What is gravitational field?A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies.
When a known mass from a spring scale is hung; by e; measuring the spring scale reading when the mass is at rest, the magnitude of the gravitational field strength ( reading/mass) can be calculated.
When a heavy metal ball is dropped, by measuring e the drop height and the time it takes the ball to hit the ground, the magnitude of the gravitational field strength ( h = gt²/2) can be calculated. Hence, option (B) and option (D) is correct.
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Question 10 (10 points)
Listen
In an ionic solution, 5.0x1015 negative ions with charge -e pass to the right each
second while 8.0x1015 positive ions with charge +2e pass to the left. What are the
magnitude and direction (+ or -) of current in the solution? (to the right is the +
direction, to the left is the - direction)
Note: Your answer is assumed to be reduced to the highest power possible.
Your Answer:
x10
Answer
units
Answer:
Please I do not understand the instructions given at the end of the question
Light containing two different wavelengths passes through a diffraction grating with 1,250 slits/cm. On a screen 17.5 cm from the grating, the third-order maximum of the shorter wavelength falls midway between the central maximum and the first side maximum for the longer wavelength. If the neighboring maxima of the longer wavelength are 8.44 mm apart on the screen, what are the wavelengths in the light
Answer:
[tex]\lambda_s =6.43*10^-4m[/tex]
Explanation:
From the question we are told that:
Diffraction grating [tex]N=1250slits/cm[/tex]
Distance b/w Screen and grating length [tex]d_{sg}=17.5 cm[/tex]
Distance b/w neighboring maxima and Screen [tex]d_{ms}=8.44[/tex]
Generally the equation for grating space is mathematically given by
[tex]d(g)=\frac{1}{N}[/tex]
[tex]d(g)=\frac{100}{1250}[/tex]
[tex]d(g)=0.08[/tex]
Generally the equation for small angle approximation is mathematically given by
[tex]\triangle y=\frac{\lambda d}{L}[/tex]
Therefore for longest wavelength
[tex]\lambda _l=\frac{8.44*10^{-3}*(0.08)}{0.175m}[/tex]
[tex]\lambda _l=3.858*10^{-3}[/tex]
Therefore the third order maximum equation for the shorter wavelength as
[tex]\lambda_s =\frac{1}{6} \lambda_l[/tex]
[tex]\lambda_s =\frac{1}{6} (3.858*10^-^3)[/tex]
[tex]\lambda_s =6.43*10^-4m[/tex]
The wavelengths in the light is given as
[tex]\lambda_s =6.43*10^-4m[/tex]
types of wave interactions include
what is the stress in a steel wire that is 5m long and 0.04cm squared in cross section If the wire bears a load of 20kg?
Answer:
Explanation:
stress = ?
length =5 m
area of cross section = 0.04 cm or 0.0004m
force = 20 × 10 = 200 N ( w = mg) g = 10
formula : stress = force / cross-sectional area
stress = 200 / 0.0004
stress = 500,000 Nm^-2
Two experiments are performed on an object to determine how much the object resists a change in its state of motion while at rest and while in motion. In the first experiment, the object is pushed with a constant known force along a horizontal surface. There is negligible friction between the surface and the object. A motion sensor is used to measure the speed of the object as it is pushed. In a second experiment, the object is tied to a string and pulled upward with a constant known force, and a motion sensor is used to measure the speed of the object as it is pulled upward. The student uses the data collected from the motion sensor to determine the mass of the object in both experiments.
Required:
What classifies the type of mass that was determined in each experiment?
Answer:
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
Explanation:
It is given that a student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass.
In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.
In the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
As per the given problem, the student performs two types of experiment to see how change in its resistance while in the state of motion and in rest.
In the first experiment, an object is pushed with a force against a horizontal surface and the speed is measured using a sensor. Here, work is done against the inertia of the object as it is pushed from rest. So the mass is inertial mass. In the second experiment, an object is pushed or thrown upwards with a force and speed is measured. Here, the mass is gravitational mass as the work done in the second experiment is against the gravity or against the weight of the object.Thus, we can conclude that the in the first experiment, the mass is inertial mass and in the second experiment, the mass is a gravitational mass.
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If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links pls
The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was
a = ∆v / ∆t = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²
By Newton's second law, the magnitude of the force F is proportional to the acceleration a according to
F = m a
where m is the object's mass. Solving for m gives
m = F / a = (10 N) / (3.52941 m/s²) ≈ 2.8 kg
Easy physics question help.!!!
Answer: This is not easy lol
Explanation:
3) Which of the following can affect the success of a start-up business?
A) the entrepreneur's ability to communicate
B) the entrepreneur's ability to resolve conflict
C) the entrepreneur's ability to work with his employees
D) all of the above
Answer:
all of the above.
Explanation:
all of these qualities are crucial in order to have a successful business.
Calculate the terminal velocity of
the following nain drops faning
through air (a) one with a diameter
of 0.3cm 6 one with a a diameter
of o. Olm. Take the density of
water to be looo Kym3 and the
eis cosity of air to be ixlos pas.
The buoyancy effect of the air
may be ignored)
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between his clothes and Earth is 0.701. He slides so that his speed is zero just as he reaches the base. (a)How much mechanical energy is lost due to friction acting on the runner
Answer:
-589.05 J
Explanation:
Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner
So, W = ΔK
W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)
So, substituting the values of the variables into the equation, we have
W = 1/2m(v₁² - v₀²)
W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)
W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)
W = 1/2 × 72.9 kg(-16.1604 m²/s²)
W = 1/2 × (-1178.09316 kgm²/s²)
W = -589.04658 kgm²/s²
W = -589.047 J
W ≅ -589.05 J
a glass bottle full of mercury has mass of 500g on being heated through 35 degree celsuis,2.43g of mercury are expelled .calculate the mass of the mercury remaining in the bottle
Answer:
The mass of the mercury remaining in the bottle is 497.57 grams.
Explanation:
The mass of the mercury remaining in the bottle is found by subtracting the mass expeled due to heating from initial mass inside the bottle. That is:
[tex]m_{f} = m_{o}-\Delta m[/tex] (1)
Where:
[tex]m_{o}[/tex] - Initial mass, in grams.
[tex]\Delta m[/tex] - Mass expelled due to heating, in grams.
[tex]m_{f}[/tex] - Final mass, in grams.
If we know that [tex]m_{o} = 500\,g[/tex] and [tex]\Delta m = 2.43\,g[/tex], then the mass of the mercury remaining in the bottle is:
[tex]m_{f} = m_{o}-\Delta m[/tex]
[tex]m_{f} = 497.57\,g[/tex]
The mass of the mercury remaining in the bottle is 497.57 grams.
A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120
seconds, what was his acceleration?
Answer:
a = 52s²
Explanation:
How to find acceleration
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
Solve
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
explain the refraction of light on a glass slab
Answer:
refraction is the change in direction of a wave passing from one medium to another or from a gradual change in the medium.
6) Which of the following describes a good team member?
A) She is willing to compromise.
B) He is aggressive.
C) She is stubborn.
D) He is conceited.
Answer: A
Explanation:
Because someone who is aggressive, stubborn, or proud of theirselves are more likely to think they're above everyone else and be a bully. However someone who is willing to compromise is better since you can generally make everyone happy that way
HOPE THIS HELPS ^^
A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to rotate around it. The box is 60 cmcm wide and 70 cmcm high. What minimum magnitude of the force F is needed to cause this movement
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N
The minimum magnitude of the force needed to cause the movement is 34.3 N.
Given:
Mass of the box (m) = 5.0 kg
Width of the box (w) = 60 cm = 0.6 m
Height of the box (h) = 70 cm = 0.7 m
The torque acting is:
τ_weight = (mg) × (h÷2)
The total torque is zero. Therefore, the force is:
F × (h÷2) + (mg) × (h÷2) = 0
F = -(mg) × (h÷2) ÷ (h÷2)
F = -(5.0 × 9.8 ) × (0.7 m ÷ 2) / (0.7 m ÷ 2)
F = -34.3 N
|F| = 34.3 N
Hence, the minimum magnitude of the force F needed to cause the movement is 34.3 N.
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A copper plate is free to swing between the poles of a large electromagnet. When the field is turned up the plate Group of answer choices will swing faster. will not be affected at all. will brake and quickly come to rest. will swing with a larger amplitude, because it is pushed by the magnet. will become a permanent magnet.
Answer:
C: will brake and quickly come to rest.
Explanation:
Correct answer is option C because for the swinging motion of the copper plates between the magnetic field which is set up as a result of it being between the two magnetic poles, there will be a continuous change of magnetic field flux that will be linked with the swinging pendulum.
As a result of this continuous change of magnetic field flux, it makes eddy currents to be set up in the copper plate which according to the Lenz's laws of electromagnetic induction tries to oppose the motion of the swinging pendulum and finally will make it come to rest.
the higher the objects " ? ", the more kinetic energy
A proton is moved so that its electric potential energy increases from 4.0 × 10-14 J to 9.0 × 10-14 J. The magnitude of the charge on a proton is 1.602 × 10-19 C.
What is the electric potential difference through which the proton moved?
2.5 × 105 V
3.1 × 105 V
5.6 × 105 V
8.1 × 105 V
Answer:
B. 3.1 × 10^5 V
Explanation:
Answer:
B
Explanation:
e2021
An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each with an average mass of 70.0 kg and it takes 20.0 s to accomplish this task, determine the mass of the elevator.
Find how much work ∆W is done by the motor in lifting the elevator:
P = ∆W / ∆t
where
• P = 45.0 kW = power provided by the motor
• ∆W = work done
• ∆t = 20.0 s = duration of time
Solve for ∆W :
∆W = P ∆t = (45.0 kW) (20.0 s) = 900 kJ
In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is M = (m + 490.0) kg, where m is the mass of the elevator alone. Then
∆W = M g h
where
• g = 9.80 m/s² = acceleration due to gravity
• h = 35.0 m = distance covered by the elevator
Solve for M, then for m :
M = ∆W / (g h) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg
m = M - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg
A car braked with a constant deceleration of 36 ft/s2, producing skid marks measuring 50 ft before coming to a stop. How fast was the car traveling when the brakes were first applied
Answer:
Initial velocity u = 60 ft/s
Explanation:
Given:
Deceleration a = -36 ft/s²
Distance covered s =50ft
Final velocity v = 0 ft/s
Find:
Initial velocity u
Computation:
Using third equation of motion;
v² = u² + 2as
0² = u² + 2(-36)(50)
0 = u² - 3600
u² = 3600
u = 60 ft/s
Initial velocity u = 60 ft/s
On the map, which major plate is flanked by the red sea rift and the Minor Arabian Plate?
A:#1 North American Plate
B:#3 South American Plate
C:#5 Eurasian Plate
D:#2 African Plate
Answer:
D:#2 African Plate
Explanation:
The African Plate is flanked by the Red sea rift and the minor African plate.
The Red sea rift is a small part of a greater line of rifts known as the Great African Rift Valley. The rift valley is making several small lakes all through Africa and it will eventually split up the African continent.
The Red sea lift is the divergent boundary between the African plate and the Arabian plate. It means that the two plates are moving apart or spreading apart.
Hi please zoom in to see it clearly, uh you don’t have to answer them all but it would be nice !!! (no links please)
Answer:
3. (a) - is the answer most likely
4: (a)
5 (d or c)
6 (b)
2. Plastic is a great conductor of charge so it moves quicker.
True
False
Answer:
the answer is false
Explanation:
plastic doesnt conduct anything
A box having a weight of 8 lb is moving around in a circle of radius rA = 2 ft with a speed of (vA)1 = 5 ft/s while connected to the end of a rope. If the rope is pulled inward with a constant speed of vr = 4 ft/s, determine the speed of the box at the instant rB = 1 ft. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box
Answer:
W = 1.875 J
Explanation:
For this exercise let's use the relationship between work and kinetic energy
W = ΔK
The kinetic energy of rotational motion is
K₀ = ½ I w²
we can assume that the box is small, so it can be treated as a point object, with moment of inertia
I = m rₐ²
angular and linear velocity are related
v = w r
w = v / r
we substitute in the equation, for point A
K₀ = ½ (m rₐ²) (v / rₐ)²
K₀ = ½ m v²
For the final point B, as the system is isolated the angular momentum is conserved
initial L₀ = Io wo
final L_f = I_f w_f
L₀ = L_f
I₀ w₀ = I_f w_f
(m rₐ²) w₀ = (m [tex]r_{b} ^2[/tex]) w_f
w_f = (rₐ/r_b)² w₀
with this value we find the final kinetic energy
K_f = ½ I_f w_f²
K_f = ½ (m [tex]r_{b}^2[/tex]) ( (rₐ / r_b)² w₀) ²
K_f = ½ m [tex]\frac{r_a^4}{r_b^2} \ w_o^2[/tex]
we substitute in the realcion of work
W = K_f - K₀
W = ½ m [tex]( \( \frac {r_a^2 }{r_b} )^2[/tex] w₀² - ½ m v²
W = ½ m [tex]\frac{r_a^4}{r_b^2} ( \frac{v}{r_a} ) ^2[/tex] - ½ m v²
W = ½ m [tex]\frac{r_a^2}{r_b^2} \ v^2[/tex] - ½ m v2
W = ½ m v² (([tex]( \ (\frac{r_a}{r_b})^2 -1)[/tex]
let's calculate
W = ½ ( [tex]\frac{8}{32}[/tex] ) 5 ((2/1)² -1)
W = 0.625 (3)
W = 1.875 J
why the speed of light decreases as it passes from air into another substance?
Answer:
If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.
which particle have a mass of 1 u
Answer:
Explanation:
proton