Calculate the time taken by the light to pass through a nucleus of diameter 1.56 10 -16 m. (speed of light is 3 10 8 m/s)

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and

natulia [17]

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\\\\\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F\times t\\\\mv=F\times t\\\\v=\dfrac{Ft}{m}\\\\v=\dfrac{120\times 0.8}{55}\\\\v=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

4 0

2 years ago

The image shows one complete cycle of a mass on a spring in simple harmonic motion. An illustration of a mass on a vertical spri

Alja [10]

Answer:

D. "The net force is zero, so the acceleration is zero"

Explanation:

edge 2020

6 0

2 years ago

A fisherman has caught a very large, 5.0kg fish from a dock that is 2.0m above the water. He is using lightweightfishing line th

agasfer [191]

Answer:

t = 2 s

Explanation:

As we know that fish is pulled upwards with uniform maximum acceleration

then we will have

$T - mg = ma$

here we know that maximum possible acceleration of so that string will not break is given as

$T = 54 N$

now we have

$54 - (5 \times 9.8) = 5 a$

$a = 1 m/s^2$

now for such acceleration we can use kinematics

$d = \frac{1}{2}at^2$

$2 = \frac{1}{2}(1) t^2$

t = 2 s

7 0

2 years ago

Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.

Elan Coil [88]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To find the variation of kinetic energy, let's use the work energy theorem

W = ΔK

∫ F .dx = K -K₀

If the body starts from rest K₀ = 0

∫ F dx cos θ = K

Since the force and displacement are in the same direction, the angle is zero, so the cosine is 1

we substitute and integrate

α ∫ x³ dx + β ∫ dx = K

α x⁴ / 4 + β x / 1 = K

we evaluate from the lower limit F = 0 to the upper limit F

α (x⁴ / 4 -0) + β (x -0) = K

K = αX⁴ / 4 + β x

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

in order to finish the calculation we must know the displacement

8 0

2 years ago

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?

spin [16.1K]

Answer:

The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip.

Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

6 0

2 years ago