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lara31 [8.8K]
3 years ago
9

NEED HELP!!!!! 10 POINTS

Physics
2 answers:
castortr0y [4]3 years ago
6 0

Answer: D.

Explanation:

satela [25.4K]3 years ago
4 0
The answer is D I’m pretty sure
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A 2.5-kg book slides horizontally and falls from a shelf 3.0 m above the floor. How much work does the force of gravity do on th
frutty [35]

Answer:

73.5 N

Explanation:

W = PE = mgh = 2.5 * 9.8 * 3.0 = 73.5 N

7 0
3 years ago
I need help with these questions please
AURORKA [14]

24.A

25.B

26.True

27. False

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7 0
4 years ago
Read 2 more answers
Can someone explain with the Answers pls?
Dovator [93]

Answer:

a) The frequency of the third harmonic is 786 Hz

b) The frequency of the first harmonic is 340 Hz

c) The frequency of the fifth harmonic is 1640 Hz

Explanation:

The rule is as follows:

If the first harmonic frequency (also called the fundamental frequency) is F, then:

The frequency of the second harmonic (also called the second overtone) is:

2*F

The frequency of the third harmonic (also called the third overtone) is:

3*F

And so on.

With this information, we can answer the questions:

a) We want to find the frequency of the third harmonic, such that the frequency of the first harmonic is 262 Hz.

Then we have F = 262Hz

And the frequency of the third harmonic will be:

3*F = 3*262Hz = 786 Hz

b) First harmonic for a string whose fifth harmonic frequency is 1700Hz.

Let's define F as the first frequency (the one we want to find)

Then the fifth harmonic frequency can be written as:

5*F = 1700Hz

With this equation we can find the value of F:

F = 1700Hz/5 = 340Hz

c) We want to find the fifth harmonic for a string whose third overtone (this is the same as the third harmonic) frequency is 984 Hz.

Then if the frequency of the first harmonic is F, we know that:

3*F = 984 Hz

With this we can find the value of F:

F = 984 Hz/3 =328 Hz

Now that we know the frequency of the first harmonic, we can find the frequency of the fifth harmonic:

5*F = 5*328 Hz = 1640 Hz

8 0
3 years ago
A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted
Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

F_c = F_g - F_N

where;

Fc is the centripetal force

F_g is downward force due to weight of the driver

F_N is upward or normal force on the drive

F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0
3 years ago
Which of the following reactions release more energy? nuclear chain reactions fusion reactions fission reactions chemical reacti
I am Lyosha [343]

Answer:

nuclear chain reactions

Explanation:

yes

6 0
4 years ago
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