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3 years ago

30 points + brainliest

Physics

2 answers:

Shtirlitz [24]3 years ago

5 0

Answer:

Explanation:

The answer is 138.75 kgm/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

It is given:

m = 0.25

F = 18,500 N = 18,500 kgm/s²

Δt = 0.0075 s

I = ?

I = F · Δt = 18,500 5kgm/s² * 0.0075 s = 18,500 * 0.0075 kgm/s² * s = 138.75 kgm/s

pishuonlain [190]3 years ago

4 0

Answer:B is the answer

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A fruit bat falls from the roof of a cave. We know that her potential energy was

bulgar [2K]

Answer:

v = 15.65 m/s

Explanation:

We use conservation of mechanical energy between initial (i) and final (f) states:

Pi + KEi = Pf + KEf

At the top of the cave at the instant the bat starts to fall, there is only potential energy since the bat's velocity is zero.

Pi = m g h = 600 J

and the KEi = 0 J (no velocity)

Knowing the height of the cave's roof (12.8 m) , we can find the mass of the bat:

m = 600 J / (g 12.5) = 4.9 kg

Using conservation of mechanical energy, the final state is:

Pf + KEf = 600 J

with Pf = 0 (just touching the ground)

KEf= 1/2 4.9 (v^2)

and we solve for the velocity:

600 J = 0 + 1/2 4.9 (v^2)

v^2 = 600 * 2 / 4.9 = 244.9

v = 15.65 m/s

5 0

2 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

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

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

Metals combine easily with nonmetals. true or false

Ad libitum [116K]

The answer for that is True.

4 0

3 years ago

What kind of model is this?

fredd [130]

Answer:

I think it's Experimental

Explanation:

it has atoms on it.

5 0

2 years ago

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How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: T

pychu [463]

Answer:

33.5 kJ

Explanation:

here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.

we have ice 100g =0.1 kg

Appplying Q=mL

Q= $0.1 kg * 335 000 J kg^{-1}$

Q = 33 500 J

Q = 33.5 kJ

3 0

3 years ago