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3 years ago

11

What space agency is predicted to be first to Mars?

1 answer:

zzz [600]3 years ago

7 0

SpaceX(NASA).

America is the greatest.

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As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen

Leto [7]

Answer:

$f_b = 29.98 Hz$

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

$f_w = f(\dfrac{c+v_r}{c+v_s})$

$f_w = 30(\dfrac{343+0}{343+6})$

$f_w = 29.48 Hz$

now frequency received by bat is equal to

$f_b = f(\dfrac{c+v_r}{c+v_s})$

$f_b = 29.48(\dfrac{343+6}{343+0})$

$f_b = 29.98 Hz$

hence the frequency hear by bat will be 29.98 Hz

7 0

3 years ago

what happens if a voltmeter is connected in series with other components of the circuit (i.e , ammeter, cell, battery, resistor

Talja [164]

Answer:the voltmeter measures the potential difference of the circuit,. Voltmeter is a device used to measure potential difference.

Explanation:

8 0

3 years ago

What happens to the particles in water as the water is heated and turns to vapor? (2 points)

Naddik [55]

Answer:

The particles will more likely to move faster since they are converted from a liquid to gas.

Rules for States of Matter:

1. Solid particles always are packed close together and don't have much space to move.

2. Liquid particles have space to move around but are still packed together, but not as close as solid.

3. Gas particles are moving freely, in fact they are in the air! Gas particles are free to move wherever. For example, the air has gas particles that are constantly bumping into each other.

Let me know if I am right =)

4 0

3 years ago

Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

Drupady [299]

Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

4 0

3 years ago

A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the

Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, $v$ , which is given by

$v=\omega r$

where $\omega$ is the angular speed and $r$ is the radius of the circular path.

$\omega$ is given by

$\omega = 2\pi f$

where $f$ is the frequency of revolution.

Thus

$v=2\pi fr$

Using values from the question,

$v=2\pi\times 1.50\times0.75$

<em>Note the conversion of 75 cm to 0.75 m</em>

$v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07$

6 0

3 years ago