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mash [69]
2 years ago
6

Which best describes what occurs when a body accelerates?

Physics
1 answer:
FinnZ [79.3K]2 years ago
6 0
A body accelerates due to net force acting on it. As newton's second law of motion states:

when a net force acts on a body, it produces acceleration in the body in the direction of the net force
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Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r
Helga [31]

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon 

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

7 0
2 years ago
Read 2 more answers
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0
2 years ago
To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to
pickupchik [31]

Answer:

specific purpose statement      

Explanation:

It is a specific purpose statement made for a persuasive speech on the question of fact.

A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.

A specific purpose statement is given to audience to persuade on specific information.

8 0
2 years ago
Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

T_{left = \frac{1}{3m}[ (805 N)( (3-1) m) + ( 395 N )( \frac{3}{2} m) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

T_{left =  \frac{1}{3m}[ 3112.5 ]

T_{left =  1037.5 N

T_{left =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

T_{Right =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

T_{Right =  2520 N - 1037.5 N

T_{Right =  1482.5 N

T_{Right =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0
1 year ago
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