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2 years ago

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An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 5

0, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?

1 answer:

rosijanka [135]2 years ago

5 0

Answer:

The momentum of the object at the end is $(3.25i+6.25j+7.5k)\ kg-m/s$

Explanation:

Given that,

Mass of object = 2.5 kg

Momentum $p= 3i+6j+7k$

Force $F=50i+50j+100k$

Time $t=5\times10^{-3}\ s$

We need to calculate the momentum of the object at the end

Using formula of impulse

$J=\Delta p$

$J=m\Delta v$ ...(I)

$J=F\times\Delta t$ ....(II)

From equation (I) and (II)

$F\times\Delta t=m\Delta v$

$F\times \Delta t=m(v_{f}-v_{i})$

$mv_{f}=F\times \Delta t+mv_{i}$

Put the value into the formula

$p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k$

$p_{f}=0.25i+0.25j+0.5k+3i+6j+7k$

$p_{f}=(3.25i+6.25j+7.5k)\ kg m/s$

Hence, The momentum of the object at the end is $(3.25i+6.25j+7.5k)\ kg-m/s$

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A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration ( $g=9.8 m/s^2$ , acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s$

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

$v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s$

So, the ball rolls off the table at 1.01 m/s.

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3 years ago

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

valina [46]

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started. So the height of the cliff is 80 m.

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3 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

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3 years ago

A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. How much energy is lost? Group of answer choices 0.5

bogdanovich [222]

Answer: A 60 g golf ball is dropped from a level of 2 m high. It rebounds to 1.5 m. Energy loss will be 0.29J

Explanation: To find the correct answer, we have to know more about the Gravitational potential energy.

<h3>What is gravitational potential energy?</h3>

The energy possessed by a body by virtue of its position in gravitational field of earth is called gravitational potential energy.
The gravitational potential energy of a body at a height h with respect to the height h will be,

$U=mgh$

Expression for gravitational potential energy loss will be,

$E=U_i-U_f$

<h3>How to solve the problem?</h3>

The total energy before the ball dropped will be,

$U_i=mgh_i=60*10^-3kg*9.8m/s^2*2m=1.176 J$

The total energy after when the ball rebounds to 1.5m will be,

$U_f=mgh_f=60*10^-3kg*9.8m/s^2*1.5m=0.882J$

The total energy loss will be,

$E=1.176-0.882=0.294J$

Thus, we can conclude that, the energy loss will be,0.294J.

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2 years ago

Salt water has greater density than fresh water. A boat floats in both fresh water and in salt 12) water. Where is the buoyant f

Vedmedyk [2.9K]

Answer:

in salt water

Explanation:

The buoyant force acting on a body depends on the volume of body immersed, density of liquid in which the body is immersed and acceleration due to gravity.

Buoyant force = Volume immersed x density of liquid x g

As the density of salt water is more than the fresh water so the buoyant force acting on the boat in salt water is more than the fresh water.

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3 years ago