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3 years ago

Which statement is true for the hang time of a projectile?

Physics

1 answer:

Andre45 [30]3 years ago

5 0

Answer:

The hang time of a projectile can be calculated solely by knowing the vertical component of the velocity of projection

Explanation:

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How much kinetic energy does a 50 kg rock has if its moving at a velocity of 2m/s

maria [59]

Answer:

KE = 100 J

Explanation: Should be correct

5 0

3 years ago

If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days?

Nat2105 [25]

The formula for the mass that remains:
$m=m_0 \times (\frac{1}{2})^\frac{t}{T}$
m₀ - the initial mass, t - time, T - the half-life

$m_0=10 \ g \\
T=14 \ d \\
t=42 \ d \\ \\
m=10 \times (\frac{1}{2})^\frac{42}{14}=10 \times (\frac{1}{2})^3=10 \times \frac{1}{8}=10 \times 0.125=1.25$

The answer is c. 1.25 g.

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3 years ago

Your friend turns the volume up on their speaker system, so that the sound intensity is 4 times greater than it was beforehand.

erica [24]

Answer:

Explanation:

We know that intensity I = P/A where P = power and A = area through which the power passes through.

Now, let the initial intensity of the speaker be I₀ and its initial power be P₀. Since the intensity is increased by a factor of 4, the new intensity be I and new power be P.

So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

E = 4E₀

Since the energy is four times the initial energy, the energy output increases by a factor of 4.

4 0

3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$

The change in momentum of the astronaut is

$\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s$

And the duration of the push is

$\Delta t = 0.600 s$

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

$F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N$

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

$K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J$

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

$K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J$

3 0

4 years ago

Commercials for a toy bouncy ball advertise that it will bounce to a height that is greater than the

storchak [24]

Answer:

because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!

7 0

3 years ago

Which statement is true for the hang time of a projectile?​

Which statement is true for the hang time of a projectile?