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1 year ago

What is the weight of an astronaut with a mass of 75 kg on the moon? The gravitational field strength on the moon is 1.6 Nkg-1.

Physics

1 answer:

bearhunter [10]1 year ago

7 0

okay this is kinda easy

<u>What is the gravitational field strength on the moon?</u>

The Moon has a gravitational field strength of 1.6 N/kg.

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An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which

katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

$\lambda=80 cm = 0.80 m$ (wavelength)

So the proper frequency is

$f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz$

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

$v_s = +35.0 m/s$ (velocity of the ambulance)

$v_o = 0$ (velocity of the staff)

Substituting,

$f'=\frac{343+0}{343+35}(429)=389 Hz$

Learn more about frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0

2 years ago

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

kifflom [539]

Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

Answer> 0.875 N*s

6 0

2 years ago

How many significant figures do each of the following numbers have: (a) 214, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 3236

Korolek [52]

In determining the number of significant figures in a given number, there are three rules to always remember / follow:

First: All integers except zero are always significant.

<span>Second: Any zeros located between non zeroes are always significant.</span>

Third: A zero located after a non zero in a decimal is always significant whether it is before or after the decimal

Therefore using this rule, the number of significant digits in the given numbers are:

(a) 214 = 3

(b) 81.60 = 4

(d) 0.03 = 1

(e) 0.0086 = 2

(f) 3236 = 4

(g) 8700 = 2

4 0

2 years ago

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

Tanya [424]

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram are:

1) gravitational pull in downward direction

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

i) F Cos 37° along the horizontal plane in forward direction and

ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 $\times$ 9.8 + 40 $\times$ 0.601 because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

3 0

2 years ago

A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless

Allisa [31]

$1.6a = \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3} \\ \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\ \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\ \\ m_3 = \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\ \\ m_3 = 22.57 kg$

4 0

2 years ago