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1 year ago

6

A ball of mass m falls vertically, hits the floor with a speed ui , and rebounds with a speed u f . what is the magnitude of the

impulse exerted on the ball by the floor

1 answer:

Sholpan [36]1 year ago

5 0

Answer:

Change of momentum = M (Vf - (-Vi)) where V represents the scalar speeds of the ball or

I = M (ui + uf) and I is the impulse ΔM V = I Force = Change in Momentum

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Question 3 (5 points)

Art [367]

Answer:

There no image

Explanation:

7 0

3 years ago

A student uses 60 Newtons of force to climb 18 meters in 6 seconds. Calculate her Power.

puteri [66]

Answer:

15

Explanation:

P=W/T

T=6sec

W=?

F=60N

S=18m

W=F X S. .s indicate displacement

W=60x18

W=108

So p=108 j/6sec

P=15watt

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2 years ago

Technician A says that you can usually depressurize a brake accumulator by turning the ignition switch ""off"" to disable the el

Rzqust [24]

Answer:

Both Technician A and Technician B

Explanation:

In order to gain a better understanding of the solution above let define some terms

Break Accumulator

We can define a break accumulator as storage that that helps generate the required pressure in order for the breaking system to respond faster this accumulator is charged by turning the steering wheel slowly at once from lock to lock now this build the pressure in the accumulator and one way to depressurize is it is by turning the ignition switch ""off""

Now a scan tool is a device that can interface with a car it can also be used to diagnose a car an get the diagnostic information to help in the cars diagnoses and also be used to reprogram a car

3 0

3 years ago

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 4.5 \times 0.13^2$

U = 0.038 J

at x = 0,the only energy will be kinetic .

$\dfrac{1}{2}mv^2=0.038$

$\dfrac{1}{2}\times 0.23 \times v^2=0.038$

v² = 0.3304

v = 0.575 m/s

displacement of the glider

using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x =v\sqrt{\dfrac{m}{k}}$

$x =3\times \sqrt{\dfrac{0.23}{4.5}}$

x = 0.678 m

8 0

2 years ago

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s

Gekata [30.6K]

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

$P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580$

7 0

2 years ago