A stone that starts at rest free falls for 7.0 s. How far does the stone fall in this time?

A wave is traveling from one medium to another.If the wave is partially transmitted an partially reflected from the boundary,wha

Kamila [148]

Answer: the answer is C

Explanation:

3 0

2 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?

12345 [234]

   Density = (mass) / (volume)

      4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by 0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

   mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

   = (360 kg) x (9.8 m/s²)

   = (360 x 9.8) kg-m/s²

   = 3,528 newtons .

That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

   (3,528 - 882) = 2,646 newtons (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.

5 0

2 years ago

Read 2 more answers

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

2 years ago

A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the collision would be totally inelastic. Since the both the vehicles locked together, they have the same final velocity.

Mass of car $m_{1}=1200 kg$

Mass of truck $m_{2}=2800 kg$

Initial speed of the car $u_{1}=100 km /h$

Initial speed of the truck $u_{2}=50 km /h$

The final velocity of the wreck will be

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

since final speed are same, $v_{1}=v_{2}=v$

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

$1200\times 100 +2800 \times 50 =(1200+2800)v\\v=\frac{260000}{4000} \\v=65 km/h$

5 0

2 years ago

You should increase the distance between your car and the vehicle ahead when you:

ELEN [110]

<span>You should increase the distance between your car and the vehicle ahead when you are being tailgated by another driver.
Tailgated is a condition when the vehicle behind you is very close to you that there is no distance between your vehicle and the vehicle behind you. In that case you should increase distance from the vehicle ahead to avoid collision in any case. These days people were mostly in hurry and everyone want to go ahead, so tailgating is very common.</span>

8 0

2 years ago