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4 years ago

5

What is the acceleration of a spaceship that is initially at rest and then speeds up to 220 m/s in 20 seconds

1 answer:

gizmo_the_mogwai [7]4 years ago

3 0

Acceleration = (change in speed) / (time for the change)

Acceleration = (220 m/s) / (20 sec)

<em>Acceleration = 11 m/s²</em>

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2. A ball is tossed straight upwards with a velocity of 60m/s. Using an estimate of gravity of -10m/s², about how many seconds w

sdas [7]

Answer: A

Explanation:

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2 years ago

Which is heavier 1kg of iron or 1 kg of feathers?

Tema [17]

-- On Earth, 1 kg of anything weighs 9.81 Newtons (2.205 pounds). It doesn't matter what substance comprises the kilogram.

-- Although their weights are identical, any mass of feathers occupies more volume than the same mass of iron, no matter where they happen to be located.

-- Iron is more dense than feathers, and more dense than a lot of other substances as well, too.

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4 years ago

Is a puddle of water an object in motion?

Brilliant_brown [7]

Yes, a puddle of water is an object in motion.

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3 years ago

. A block of mass 4.0 kg on a spring has displacement as a function of time given by, x (t) = (1.0 cm) cos [(2.0 rad/s) t + 0.25

svlad2 [7]

Answer and Explanation:

1. Evaluate the function x(t) at t=0.5

$x(t)=1cos(2t+0.25)\\x(0.5)=cos(2*0.5+0.25)=cos(1+0.25)=cos(1.25)=0.3153223624\approx0.31cm$

2. The period of motion T can be calculated as:

$T=\frac{2\pi}{\omega}$

Where:

$\omega=2rad/s$

So:

$T=\frac{2\pi}{2}=\pi\approx3.14s$

3. The angular frequency can be expressed as:

$\omega=\sqrt{\frac{k}{m} }$

Solving for k:

$k=(\omega)^2*m=(2)^2*4=4*4=16\frac{N}{m}$

4. Find the derivate of x(t):

$\frac{dx}{dt} =v(t)=-2sin(2t+0.25)$

Now, the sine function reach its maximum value at π/2 so:

$2t+025=\frac{\pi}{2}$

Solving for t:

$t=\frac{\frac{\pi}{2} -0.25}{2} =0.6603981634s$

Evaluating v(t) for 0.6603981634:

$v(0.6603981634)=-2sin(2*0.6603981634+0.25)=-2sin(\frac{\pi}{2} )=-2*1=2$

So the maximum speed of the block is:

$v(0.6603981634)=2cm/s=0.02m/s$

In the negative direction of x-axis

5. The force is given by:

$F=kx$

The cosine function reach its maximum value at 2π so:

$2t+0.25=2\pi$

Solving for t:

$t=\frac{2\pi-0.25}{2} =3.016592654s$

Evaluating x(t) for 3.016592654:

$x(3.016592654)=cos(2*3.016592654+0.25)=cos(2\pi)=1cm=0.01m$

Therefore the the maximum force on the block is:

$F=16*0.01=0.16N$

6 0

3 years ago

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid

Setler79 [48]

Answer:

a)

$- 7.04\times10^{11} J$

b)

$5.03\times10^{5} N$

Explanation:

a)

$m$ = mass of the asteroid = 43000 kg

$v_{o}$ = initial speed of asteroid = 7600 m/s

$v$ = final speed of asteroid = 5000 m/s

$W$ = Work done by the force on asteroid

Using work-change in kinetic energy theorem

$W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J$

b)

$F$ = magnitude of force on asteroid

$d$ = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

$W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N$

4 0

3 years ago