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3 years ago

8

A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward.

If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.

2 answers:

marusya05 [52]3 years ago

6 0

Answer:

Force on superball will be $=9.8222\times 10^{-4}N$

Explanation:

We have given mass of superball m = 52 gram = 0.052 kg

Velocity change from 20 m/sec downward to 14 m/sec upward

Let downward velocity is positive then upward velocity is negative

So downward velocity is + 20 m/sec and upward velocity is -14 m/sec

Time is given as 1800 sec

We know that acceleration is rate of change of velocity

So $a=\frac{20-(-14)}{1800}=0.0188m/sec^2$

According to newton second law

Force = ma = 0.052×0.0188 $=9.8222\times 10^{-4}N$

Svetlanka [38]3 years ago

6 0

Answer:

0.982 N

Explanation:

mass of ball, m = 52 g = 0.052 kg

initial velocity, u = - 20 m/s (downward)

final velocity, v = 14 m/s (upward)

time of contact, t = 1.8 s

According to Newton's second law, the rate of change of momentum of the body is equal to the forced exerted on that body.

initial momentum , pi = mass x initial velocity = 0.052 x (-20) = - 1.04 kg m/s

final momentum, pf = mass x final velocity = 0.052 x 14 = 0.728 kg m/s

Change in momentum = final momentum - initial momentum

= 0.728 - (- 1.04)= 1.768 kg m/s

So, force = change in momentum / time

Force = 1.768 / 1.8 = 0.982 N

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The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to

DedPeter [7]

Answer:

$9.4\cdot 10^{10} m$

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

$\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}$

where

$r_o$ is the distance of the new object from the sun (orbital radius)

$T_o=180 d$ is the orbital period of the object

$r_e = 1.50\cdot 10^{11} m$ is the orbital radius of the Earth

$T_e=365 d$ is the orbital period the Earth

Solving the equation for $r_o$ , we find

$r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m$

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3 years ago

a train leaves a station heading east at 50 mph from that same station a car drives north at a speed of 30mph after 3 hours how

Alecsey [184]

Answer:

d = 175 miles

Explanation:

Train is moving towards East with constant speed of 50 mph

While car is moving at speed of 30 mph

so after t = 3 hours

the distance moved by the train is given as

$d_1 = vt$

$d_1 = (50)(3) = 150 miles$

at the same time the distance moved by the car is given as

$d_2 = 30(3) = 90 miles$

now we know that both car and train is moving perpendicular to each other

So the distance between them after t = 3 hours is given as

$d= \sqrt{d_1^2 + d_2^2}$

$d = \sqrt{150^2 + 90^2}$

$d = 175 miles$

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3 years ago

The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t

KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is $F_f = \mu_k N$ = 0.2 N

$+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}$

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

$\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}$

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

$\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft$

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

$\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}$

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

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