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2 years ago

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A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha

t is the impulse given to the ball?

1 answer:

Diano4ka-milaya [45]2 years ago

3 0

To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

$p = m\Delta v$

Here,

m = mass

$\Delta v =$ Change in velocity

As we can see there are two types of velocity at the moment the object makes the impact,

the first would be the initial velocity perpendicular to the wall and the final velocity perpendicular to the wall.

That is to say,

$v_i = vcos\theta$

$v_f = -v sin\theta$

El angulo dado es de 45° y la velocidad de 25, por tanto

$v_i = (25)cos(45) = 17.68m/s$

$v_f = -(25)sin(45) = -17.68m/s$

The change of sign indicates a change in the direction of the object.

Therefore the impulse would be as

$p = 0.060(-17.68-17.68)$

$p = -2.12kg \cdot m/s$

The negative sign indicates that the pulse is in the opposite direction of the initial velocity.

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A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as he

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Answer:

n = 1.89 x 10¹⁹ photons/s

Explanation:

given,

Power of bulb = 100 W

Visible light = 5 W

wavelength of the visible light = 600 nm

number of photons emitted per second

using formula of wavelength

$\lambda = \dfrac{c}{\nu}$

$\nu= \dfrac{c}{\lambda}$

$\nu= \dfrac{3 \times 10^8}{600 \times 10^{-9}}$

$\nu=5 \times 10^{14}\ Hz$

energy of photon

U = h υ

U = 6.63 x 10⁻³⁴ x 4 x 10¹⁴

U = 2.65 x 10⁻¹⁹ J

number of photon

$n = \dfrac{P}{U}$

$n = \dfrac{5}{2.65 \times 10^{-19}}$

n = 1.89 x 10¹⁹ photons/s

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2 years ago

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

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2 years ago

To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch

sergij07 [2.7K]

Answer:

man upward acceleration is 0.14m/s^2

Explanation:

given data:

mass of man = 72 kg

downward force = 360 N

The mass of man of weight 72 kg is hang from two sections of rope, one section pf rope ties around man waist and other section is ties in man hands. when he pulls down the rope with 360 N force then each section of rope pulls with 360 N

we know that

Weight= mass × gravity= 72kg × 9.8 = 705.6N

Force = mass× acceleration

Force= -705.6 + (2 × 358) = 10.4 N

$acceleration = \frac{10.4}{72} = 0.14m/s^2$

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2 years ago

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

A rope is attached to a block. The rope pulls on the block with a force of 240 N, at an angle of 40 degrees to the horizontal (t

Tom [10]

Answer:

X Component is 183.85N

Explanation:

The x component of the force on the block due to the rope;

X = F cos @ where if is the force, @ is the angle mad with the block.

X = F cos @

X = 240 cos 40

Cos 40= 0.7660, so

X = 240 × 0.7660

X component= 183.85N// rounded to two decimal places.

3 0

2 years ago

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