Match each situation below to a letter on the illustration.

A rocket of mass 40000kg propelled by a force of 10^6N acquires a speed of 3000m/s determine the power expended

Katyanochek1 [597]

Answer:

3×10⁹ W

Explanation:

Power = work / time

Power = force × distance / time

Power = force × velocity

P = Fv

P = (10⁶ N) (3000 m/s)

P = 3×10⁹ W

6 0

3 years ago

Convection requires______to transfer heat.

mylen [45]

The movement of fluid

6 0

3 years ago

A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j

sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

Mechanical energy lost is equal top gain in kinetic energy

$ME_{Lost}=\dfrac{1}{2}mv^2$

$ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2$

$ME_{Lost}=662.29\ J$

b) Work done = Force x displacement

W = F. x

F = μ mg

W = μ mg . x

Work done is equal to 662.29 J

$x=\dfrac{W}{\mu m g}$

using the coefficient of the friction,μ = 0.7

$x=\dfrac{662.29}{ 0.7\times 70\times 9.8}$

x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0

3 years ago

in a certain pinhole camera, the screen is 10 cm from the pinhole. When the pinhole is placed 6 cm away from a tree ,a sharp ima

vagabundo [1.1K]

Answer:

Height of the tree is <u>9.6 cm</u>

Explanation:

We know that, Magnification of an image is written as follows.

$\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)$

Where,

$\begin{array}{l}{\mathrm{H}_{0}=\text { height of the object }} \\ {\mathrm{H}_{\mathrm{i}}=\text { height of the image }} \\ {\mathrm{D}_{0}=\text { distance of the object }} \\ {\mathrm{D}_{\mathrm{i}}=\text { distance of the image }}\end{array}$

As per given question,

$\begin{array}{l}{\mathrm{H}_{1}=\text { height of the image }=\text { height of the image of the tree on screen }=16 \mathrm{cm}} \\ {\mathrm{D}_{0}=\text { distance of the object }=\text { distance of the tree from the pinhole }=6 \mathrm{cm}} \\ {D_{1}=\text { distance of the image }=\text { distance of the image from the pinhole }=10 \mathrm{cm}} \\ {\mathrm{H}_{0}=\text { height of the object }=\text { height of the tree }}\end{array}$

Substitute the values in the above formula,

$\begin{array}{l}{\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)} \\ {\left(\frac{16}{H_{o}}\right)=\left(\frac{10}{6}\right)} \\ {\mathrm{H}_{\mathrm{o}}=\left(\frac{16 \times 6}{10}\right)} \\ {\mathrm{H}_{\mathrm{o}}=9.6 \mathrm{cm}}\end{array}$

Height of the tree is 9.6 cm.

8 0

3 years ago

What is the main difference between a maglev train and a normal train?

professor190 [17]

Regular trains follow train tracks and travel on rails; Maglev trains are Magnetically Levitated using powerful electromagnets, so that they actually hover slightly above the ground. This massively reduces friction and allows them to travel much faster (kind of like how a puck can go really far with only a little push on an air hockey table).

5 0

3 years ago