Ask question

Ask question

All categories

3 years ago

6

A hot-air balloon with a mass of 400 kilograms moves across the sky with 3,200 joules of kinetic energy. The velocity of the bal

loon is meters/second.

1 answer:

kotegsom [21]3 years ago

4 0

Answer:

4 m/s

Explanation:

KE =

Velocity of balloon will be 4 m/s.

!! Hope It Helps !!

You might be interested in

On a cloudless day, the sunlight that reaches the surface of the earth has an intensity of about W/m². What is the electromagnet

Otrada [13]

Answer:

1.907 x 10⁻⁵ J.

Explanation:

Given,

Volume of space, V = 5.20 m³

Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².

Electromagnetic energy = ?

$E = \mu V$

$E = (\dfrac{S}{c})\times V$

where c is the speed of light.

$E = (\dfrac{1.1\times 10^3}{3\times 10^8})\times 5.20$

$E = 1.907\times 10^{-5}\ J$

Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.

4 0

4 years ago

Air is composed of many gases, while water is composed of a pure substance. Which best compares air and water?

natta225 [31]

Answer: c

Explanation:

C Air is a compound of two or more components that keep their own identifying properties, while water is composed of mixtures that combine to form a compound.

4 0

3 years ago

Read 2 more answers

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

3 0

4 years ago

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

3 years ago

The answer and how to do it

Arturiano [62]

Current = charge per second
2 Coulombs per second = 2 Amperes

Potential difference = (current)x(resistance) in volts.

That's (2 Amperes) x (2 ohms).

That's how to do it.
I think you can find the answer now.

8 0

4 years ago