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3 years ago

Can anyone help me with this assignment? please and thank you

Physics

1 answer:

Vikki [24]3 years ago

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I’m not able to see it

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A bus initially moving at 20 m/s with an acceleration of -4m/s² for 5

Alborosie

Answer:

50m; 0m/s.

Explanation:

Given the following data;

Initial velocity = 20m/s

Acceleration, a = - 4m/s²

Time, t = 5secs

To find the displacement, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting into the equation, we have;

$S =20*5 + \frac{1}{2}*(-4)*5^{2}$

$S =100 + (-2)*25$

$S =100 - 50$

S = 50m

Next, to find the final velocity, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

<em>Substituting into the equation, we have;</em>

$V^{2} = 20^{2} + 2(-4)*50$

$V^{2} = 400 - 400$

$V^{2} = 0$

V = 0m/s

<em>Therefore, the displacement of the bus is 50m and its final velocity is 0m/s.</em>

5 0

3 years ago

A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

Ganezh [65]

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

4 0

3 years ago

. Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.

anastassius [24]

Ke = (1/2)mv²

m = 100kg, v = 10 km/s = 10*1000 = 10000m/s

Ke = (1/2)*100*10000

Ke = 500000 Joules

7 0

3 years ago

Read 2 more answers

Two resistors A and B are arranged in series in one branch of a parallel arrangement. The other branch contains a single resisto

Ganezh [65]

Answer:

Explanation:

A and B are in series , Total resistance = Ra + Rb

This resistance is in parallel with single resistor C

Equivalent resistance Re = Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )]

Now this combination is in series in single resistance D .

Total resistance = Rd + Re

= Rd + { Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )] }

5 0

2 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago