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2 years ago

Acceleration is a measure of the change in the velocity of an object per unit time true or false

Physics

2 answers:

snow_lady [41]2 years ago

8 0

this is true...acc= change in velocity / time

Ksju [112]2 years ago

3 0

This is true. Hope I helped!

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Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

spin [16.1K]

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

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2 years ago

After rubbing a balloon on your shirt, your hair sticks up when the balloon is near your head. Explain how tiny particles on you

skad [1K]

Almost right. protons are positive and electrons are negative. so when you run the balloon on your hair, electrons are transferred between them (i’m not sure which direction) and now one is positively charged as it lost negative particles and one is negative as it gained negative particles (electrons). opposite charges attract

3 0

1 year 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:

$F \Delta t = \Delta p$