When a tennis player practices by

A rope exerts a force F on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of t

tia_tia [17]

Answer:

400 J

Explanation:

Given:

Δy = 4.00 m

v₀ = 0 m/s

a = 5.00 m/s²

Find: v²

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (5.00 m/s²) (4.00 m)

v² = 40.0 m²/s²

Find KE:

KE = ½ mv²

KE = ½ (20.0 kg) (40.0 m²/s²)

KE = 400 J

5 0

3 years ago

One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sampl

Radda [10]

Answer:

E = 1.38 x 10⁸ eV = 138 MeV

Explanation:

The energy associated with the given wavelength can be found from the following formula:

$E = \frac{hc}{\lambda}$

where,

E = Energy of electron = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = Speed of Light = 3 x 10⁸ m/s

λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m

Therefore,

$E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})$

5 0

3 years ago

Your car is skidding to a stop from a high speed. name all the forces that apply

DerKrebs [107]

Normal force, weight, Kinetic friction, and air resistance are a few I think of the top of my head.
I hope this helps

6 0

3 years ago

The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of frict

blondinia [14]

Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

$T - m*g = m*a\\\\$ ... Eq 1

Where,

a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

$N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )$

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

$F = u*N\\\\F = u*M*g *cos ( Q )$

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

$M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\$ .. Eq2

- Add the two equation, Eq 1 and Eq 2:

$M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}$

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

$T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N$

4 0

3 years ago

Which of the following may have happened if the inner planets had greater masses during the formation of the solar system? . . A

Svetach [21]

Right answer is option b that is the inner planet may havebeen larger and more gaseous.

5 0

3 years ago