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sergiy2304 [10]
3 years ago
8

If the motor M rotates in the direction shown by the arrow as illustrated in the diagram below, what is going on? A. 1 and 2 are

going down B. 1 goes up and 2 goes down C. 1 and 2 are going up D. 1 goes down and 2 goes up 49) For the system in the diagram below to be in equilibrium, what is the mass of P in pounds? The mass at the other end of the beam is 5 kg.
Physics
1 answer:
madreJ [45]3 years ago
7 0

Answer:

C. 1 and 2 are going up

Explanation:

The motor turns counterclockwise, so the pulley above it will also turn counterclockwise.  This means the pulley to the right will turn counterclockwise.  So block 2 will move up.

Since the belt is twisted on the left side, the pulley on the left will spin in the opposite direction (clockwise).  So block 1 will also move up.

You might be interested in
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

6 0
4 years ago
A 1000 kg car accelerates from rest at a rate of 10 m/s² for 3 seconds. A) what is the final velocity of the car?
Lorico [155]

Answer:

Refer to the attachment!~

5 0
3 years ago
What is the change in internal energy if 60J of heat are released from a system and 20J of work is done on the system? Use U=Q-W
Ganezh [65]
D because 60-20= 40J
5 0
3 years ago
We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2
monitta

Answer:

The number of atoms is N = 4.37*10^{22} \ atoms

Explanation:

From the question we are told that

                 The mass of coin  is m_n = 4.2g

                   Number of atom in one mole = n =6.02*10^{23} \ atoms

                   Molar mass of nickel M = 57.8g

Now the relation to obtain the number of atom in  the  nickel coin is

                        N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel }  * No\ of\atoms \ in \ \ one\  mole\ of\ nickel

                           = \frac{4.2}{57.8}* 6.02*10^{23}

                           =4.37 *10^{22} atoms

                 

     

8 0
3 years ago
9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
Korolek [52]

<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

=686/544.8

=1.259ms^2

4 0
3 years ago
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