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algol [13]
3 years ago
15

Ball B is suspended from a cord of length l attached to cart A, which can roll freely on a frictionless, horizontal track. The b

all and the cart have the same mass m. If the cart is given an initial horizontal velocity v0 while the ball is at rest, describe the subsequent motion of the system, specifying the velocities of A and B for the following successive values of the angle θ (assume positive counterclockwise) that the cord will form with the vertical:
(a) θ = θmax
(b) θ = 0
(c) θ = θmin
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

a.Velocity of A and Velocity of B will be opposite in direction.

b.Both velocities have the same direction.

c. The two objects will move in a direction of some angle α.  

Explanation:

When the angle gets maximum the velocities of the two objects pull each other and move in opposite direction. But when the two objects make an angle zero degree with each other their motion is in the same direction.  

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The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-
nikdorinn [45]

Answer:

The answer is "60.74^{\circ}".

Explanation:

Cavity and benzene should be extended in equal quantities.

\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to  (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\

\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\

\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\

5 0
3 years ago
a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

-Paper airplane will glide to the ground do to its wings and will hit the ground last

3 0
3 years ago
Read 2 more answers
What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest
lina2011 [118]

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed  =  (27 - 0)  =  27 m/s
Time for the change = 10 s

Magnitude of acceleration =  (27 m/s) / (10 s)  =  2.7 m/s²  .
4 0
3 years ago
A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have
denpristay [2]
1) The total mechanical energy of the rock is:
E=U+K
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
E_i=U=mgh
where m=4 kg is the mass, g=9.81 m/s^2 is the gravitational acceleration and h=5 m is the height.
Putting the numbers in, we find the potential energy
U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
E_f=K= \frac{1}{2}mv^2
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
K=U=196.2 J

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
W=196.2 J-0 J=196.2 J
6 0
3 years ago
Convert 0.700 atm of pressure to its equivalent in millimeters of mercury.
inna [77]

Answer:

532 millimeters of mercury

Explanation:

In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:

1 atm = 760 mm Hg

Therefore, we can solve the problem by setting up the following proportion:

1 atm : 760 mmHg = 0.700 atm : x

Solving for x, we find

x=\frac{(760 mmHg)(0.700 atm)}{1 atm}=532 mmHg

5 0
3 years ago
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