The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.

Answer:

India is a peninsula.

Explanation:

India is called as Indian Peninsula because it is surrounded by the Indian ocean on the south, the Arabian sea on the west and the Bay of Bengal on the east.

Answer:

Coefficient of kinetic friction = 0.146

Explanation:

Given:

Mass of sled (m) = 18 kg

Horizontal force (F) = 30 N

FInal speed (v) = 2 m/s

Distance (s) = 8.5 m

Find:

Coefficient of kinetic friction.

Computation:

Initial speed (u) = 0 m/s

v² - u² = 2as

2(8.5)a = 2² - 0²

a = 0.2352 m/s²

Nweton's law of :

F (net) = ma

30N - μf = 18 (0.2352)

30 - 4.2336 = μ(mg)

25.7664 =  μ(18)(9.8)

μ = 0.146

Coefficient of kinetic friction = 0.146

Answer:

24 Newtons

Explanation:

The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.

The mass of those two cubes combined is 6 + 6 = 12 kg

So, using the following equation, we can find the force:

Force = mass * acceleration

Force = 12 * 2

Force = 24 Newtons

Answer:

A= The type of plant

B= How tall the plant is

Explanation:

Answer:

hypothesis , hope it helps

Explanation:

Answer:

Inference

Explanation:

Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.

Answer:

EACH SEX CELL WILL HAVE 10 CHROMOSOMES BECAUSE n+n=2n

means haploid parent cells join or fuse to form diploid zygote

Answer:

10

Explanation:

Answer:First option

Explanation:

hope it helped

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .

Answer:

Explanation:

From Newton's second Law of Motion,

F = ma

Ff + F = ma

Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.

Magnitude of the acceleration:

a = Ff+F/m

This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive

Answer:

17.85°

Explanation:

To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]

n1: index of refraction of Sophia's cornea = 1.387

n2: index of refraction of aqueous fluid = 1.36

θ1: angle to normal in the first medium = 17.5°

θ2: angle to normal in the second medium

You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:

[tex]\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°[/tex]

hence, the angle to normal in the aqueous medium is 17.85°

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  B

Explanation:

The system is best considered a closed system because looking at process we can see that there was no exchange of matter between the system and the surrounding,(as the was no escape of matter from the system to the surrounding )

Secondly we can  deduce that there is  a variation in the volume. from [tex]1.0 m^3[/tex] to [tex]0.5 m^3[/tex]

Answer: Around 0:35 Pm or 12:35 Am

Explanation:

The equation that describes the cooling of objects can be written as:

T(t) = Ta + (Ti - Ta)*e^(k*t)

Where Ta is the ambient temperature, here Ta = 25°C.

Ti is the initial temperature of the body, we have Ti = 37°C.

t is the time.

k is a constant.

So our equation is:

T(t) = 25°C +12°C*e^(k*t)

at 2pm, the temperature was 33°C

at 3pm, the temperature was 31°C.

we want to find the hour where we have our t = 0, suppose this hour is X.

then we can write our times as:

2pm ---> 2 - X

3pm ----> 3 - X

and our equations are:

33°C = 25°C + 12°C*e^(k2 - k*X)

31° = 25°C + 12°C*e^(k3 - k*X)

So we have two equations and two variables, let's solve the system.

first, simplify it a bit, for the first eq:

33 - 25 = 12*e^(k2 - k*X)

8/12 = e^(k2 - k*X)

ln(8/12) = k*2 - k*X

for the second equation we have:

31 - 25 = 12*e^(k3 - k*X)

6/12 = e^(k3 - k*X)

ln(6/12) = k*3 - k*X

So our equations are:

1) ln(2/3) = 2*k - X*k

2) ln(1/2) = 3*k - X*k

First, let's isolate one of the variables in one of the equations. let's isolate k in the first equation.

ln(2/3)/(2-X) = k

now we can replace it in the second equation:

ln(1/2) = 3*ln(2/3)/(2 - X) - X*ln(2/3)/(2-X)

now let's solve it for X, i will take a = ln(1/2) and b = ln(2/3) so it is easier to read.

a = 3*b/(2 - X) - X*b/(2 - X)

a*(2 - X) = 3*b - X*b

2a - aX = 3b - Xb

X(a - b) = 2a - 3b

X = (2*ln(1/2) - 3*ln(2/3))/(ln(1/2) - ln(2/3)) = 0.590

now, knowing that one hour has 60 minutes, then this is:

0.59*60m = 35 minutes

So the hour of death is 0:35 Pm or 12:35 Am

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

Answer:

Ff = 33.4N

Explanation:

To find the frictional force while the object is moving, you take into account that the friction force depends of the coefficient of kinetic friction.

The frictional force is given by:

[tex]F_f=\mu_kN[/tex]         (1)

Ff: frictional force = ?

µk: coefficient of kinetic friction = 0.167

N: normal force of the object = 200N

You replace the values of the parameters in the equation (1):

[tex]F_f=(0.167)(200N)=33.4N[/tex]

The frictional force, while the objects is moving, is 33.4N

Answer:

Angle = 18.41°

Explanation:

Torque = F•r•sin θ

where;

F = force

r = distance from the rotation point

θ = the angle between the force and the radius vector.

We are given;

Torque = 15 N.m

F = 95 N

r = 0.5 m

Thus, plugging in the relevant values ;

15 = 95 × 0.5 × sin θ

sin θ = 15/(95 × 0.5)

sin θ = 0.3158

θ = sin^(-1)0.3158

θ = 18.41°

Answer:

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Explanation:

Relative velocity with respect to the ground is;

Vbg = velocity of train with respect to the ground + your velocity with respect to the train

Vbg = Vtg + Vbt ......1

Given;

velocity of train with respect to the ground;

Vtg = 19 m/s

your velocity with respect to the train;

Vbt = 1.5 m/s

Substituting the given values into the equation 1;

Vbg = 19 m/s + 1.5 m/s

Vbg = 20.5 m/s

your velocity with respect to the ground Vbg = 20.5 m/s

Answer:

1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation

Explanation:

The kinetic  friction works against the kinetic energy of the car and the car stops when these two equalises .

friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.

= μk x mg

work done by friction

= force x displacement

=  μk x mg x d

kinetic energy of car at the time of accident = 1/2 m v²

kinetic energy = work done by friction

1/2 m v² = μk x mg x d

d = v² / (2 μk x g)

v² = 2dμk g

v = √(2dμk g)

Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation

Answer:

a)  v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

For the boy

         y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

        y = ½ g t2

For superman

        y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

         t = √ (2 y₀ / g)

         t = √ (2 180 / 9,8)

         t = 6.06 s

in the equation for superman, we clear the volume and calculate

         v₀ (t-5) = -y₀ + ½ g (t-5)²

         v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

         v₀ 1.06 = -174.49

         v₀ = - 174.49 / 1.06

         v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

  t (s)      Y_boy (m)   Y_superman (m)

    0             180                 180

   1              175.1               180

   5              57.5              180

   6                3.6                10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

           y = -1/2 g t²

           y = ½ 9.8 5²

           y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

Answer:

[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Explanation:

Angular acceleration

[tex]\begin{aligned}

\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\

\omega_{i} &=0 \\

\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\

&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\

&=27.02 \mathrm{rad} / \mathrm{s} \\

\alpha &=\frac{(27.02-0)}{3.15} \\

&=8.57 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

a)Tangential acceleration

[tex]\begin{aligned}

a &=r \alpha \\

&=\frac{12}{2} \times 10^{-2} \times 8.57 \\

a &=0.51 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

This question involves the concepts of the equations of motion for angular motion.

The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".

First, we will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 3.05 s

Therefore,

[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]

Now, the tangential acceleration can be given as follows:

[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]

a = 0.532 m/s²

Learn more about the angular motion here:

brainly.com/question/14979994?referrer=searchResults

The attached picture shows the angular equations of motion.

Answer:

The distance is  [tex]S = 39.2 \ m[/tex]

Explanation:

From the question we are told that

    The distance covered after t = 1 s is  [tex]d = 4.9 \ m[/tex]

According to the equation of motion

      [tex]v^2 = u^2 + 2ad[/tex]

 Now  u  =  0 m/s  since before the drop the ball was at rest

     [tex]v^2 = 2ad[/tex]

here  [tex]a =g = 9.8 \ m/s^2[/tex]

    So

       [tex]v = 9.8 m/s[/tex]

Also from equation of motion we have that

     [tex]S = ut + \frac{1}{2} at^2[/tex]

Now at  t = 2 s , as given from the question

  Then  u =  v = 9.8 m/s

And

     [tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]

     [tex]S = 9.8 * 2 + \frac{1}{2} * (9.8) * (2^2)[/tex]

    [tex]S = 39.2 \ m[/tex]

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

[tex]\int\limits^{10}_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)[/tex]

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

Answer:

0.00000009512

Explanation:

Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.

let's pass the number 9,512 10⁻⁸ to decimal notation

       9,512 / 10⁸ = 9,512 / 100000000

        0.00000009512

As we see writing this number, it is very easy to make mistakes

Answer:

Plutonium–238

Explanation:

The stability of isotopes is largely dependent on their half-life.

Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.

From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.

We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.

Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.

Answer:

d

Explanation:

good question. now the bus is moving in constant velocity . a student in front tosses a ball to the student in back. but we dont know the speed at which the student tosses a ball. we have to assume the speed

assume the speed of ball is slightly less than the speed of bus. in this case the stationary observer sees the ball in slower speed than the one inside the bus.

so a is correct

now assume the speed of ball is 1/2 the speed of bus. here stationary observer sees the ball the same speed as the one in bus observe

b is correct

assume the speed of ball is very small than the speed of bus . in this case the stationary observer see in grater speed than the student in bus

e also correct

so correct answer is d. it depends on the speed of ball tossed by the student in front.

Answer:

a) Fs = (μs*mp*g)/2  |  b) τ = Fs*lp  |  c) Wτ,constant = τΘ

Explanation:

a) Fs = (μs*mp*g)/2

b) τ = Fs*lp

c) Wτ,constant = τΘ

ANSWER: num 1 is black hole

Explanation:

Complete the first and second sentences, choosing the correct answer from the given ones.

1. T = 100 K

[tex]^{\circ}C=K-273[/tex]

Put T = 100 K

[tex]T=100-273=-173^{\circ} C[/tex]

A temperature of 100 K corresponds on a Celsius scale to (-173 °C)

2. T = 50 °C

[tex]K=^{\circ}C+273\\\\K=50+273\\\\T=323\ K[/tex]

So, At 50 °C, it corresponds to a Kelvin scale of 323 K.

Answer:

Answer D is correct

Explanation:

[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]

hope this helps

brainliest appreciated

good luck! have a nice day!

Answer:

The lifetime of the particle is  [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Explanation:

From the question we are told that

    The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]

    The intrinsic width is  [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]

The lifetime is mathematically represented as

     [tex]\Delta t = \frac{h}{\Delta E}[/tex]

Where h is the Planck's constant with a value of  [tex]1.055*10^{-34} \ J\cdot s[/tex]  

substituting values

    [tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]

     [tex]\Delta t = 2.6*10^{-25} \ s[/tex]