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3 years ago

14

I need help with this Physics problem. I've been stuck forever:

1 answer:

snow_tiger [21]3 years ago

6 0

<span>Ball A with a mass of 0.500 kg is moving east at a velocity of 0.800 m/s. It strikes ball B, also of mass 0.500 kg, which is stationary. Ball A glances off B at an angle of 40.0° north of its...Two smart cars depart from the same starting location at the same time and travel different routes to the same destination, arriving at the same time. Explain why the cars travelled different...Given electric flux density D=0.3r^2 ar nc/m^2 in free space a. Find the total charge within the sphere r=3 b. Find the total electric flux leaving the sphere r=4</span>

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hydroelectric energy is based on the blank energy of water being converted to blank energy as waterfalls and pushes against the

Naddika [18.5K]

This is an example of kinetic to potential because the water is moving, pushing a turbine (Kinetic) and the turbine converts the energy to potential energy.

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3 years ago

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Calculate the density of each ball. Use the formula

Paraphin [41]

Answer:

⇒ 0.07 g/cm3

⇒ 1.15 g/cm3

Explanation:

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A force of 13 N is used to slide a 0.8 kg book across a table. What is the magnitude of the book’s acceleration?

Mkey [24]

Explanation:

acceleration=F/m

a=13/0.8

a=16 25m/s²

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4 years ago

A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

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

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

$W=mg$

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

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3 years ago

The precision of a balance is ±0.02 g. The accepted value for a measurement is 26.86 g. Which measurement is in the accepted ran

HACTEHA [7]

Answer:

Ben türküm sizi anlamıyom kb

hepiniz malsınız amk

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