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Alex Ar [27]
2 years ago
11

Two students have fitted their scooters with the same engine. Student A and his

Physics
1 answer:
sammy [17]2 years ago
3 0

The force exerted by student A with his scooter is 306 N and that of student B is 204 N.

<h3>Force applied by each student</h3>

The force exerted by each student is calculated from Newton's second law of motion.

F = ma

where;

  • m is mass
  • a is acceleration

F(A) = 127.5 x 2.4

F(A) = 306 N

F(B) = 120 x 1.7

F(B) = 204 N

Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

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What must you know to find the amount of work done on an object
Cerrena [4.2K]

Answer:

The work is calculated by multiplying the force by the amount of movement of an object (W = F * d). A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy – joules.

Explanation:

5 0
3 years ago
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2) If the initial velocity of an object is equal to a final velocity, what is the acceleration of the object?
Arlecino [84]

Answer:

The acceleration is 0.

Explanation:

The formula for acceleration is:

a = (v-u)/t

If v and u are equal, thus it would be 0 when subtracted, and anything divided by 0 is 0.

8 0
3 years ago
The drawing shows a large cube (mass = 21.0 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MaRussiya [10]

Answer:

The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.

Explanation:

The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.

The large cube in red is being pushed from the left by a force \vec{P} whose value is to be found. That cube has its own weight \vec{w}_1=m_1\vec{g}, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force \vec{N}_1.

A second cube is being pushed by the first, and since the force \vec{P} is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight \vec{w}_2=m_2\vec{g} pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as Fr=\mu~N, where \mu is the coefficient of static friction between the cubes.

Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.

In the vertical component: \Sigma F^2_y=Fr-w_2=m_2 a_y Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero a_y=0, and making explicit the other forces: \mu N_2-m_2g=0\quad\Rightarrow (0.710)N_2-(4.5)(10)=0\quad\Rightarrow N_2=(4.5)(10)/(0.710)\approx 63.38 [N]. In the last equation gravity's acceleration was rounded to 10 [m/s^2].

In its horizontal component: \Sigma F^2_x=N_2=m_2 a_x, this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined: 63.38=(4.5) a_x \quad \Rightarrow a_x=(63.38)/(4.5)\approx 14.08 [m/s^2]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.

Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: \Sigma F_x=P=m a_x, since the force \vec{P} is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together; P=(21.0+4.5) 14.08\approx 359.04 [N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.

4 0
3 years ago
You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po
nataly862011 [7]

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

4 0
3 years ago
What is the power of a hair dryer if it uses 72,000 joules energy in 60.0 seconds
Vladimir [108]
Power=Energy
72,000/60 = 1200 watts 
:)
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3 years ago
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