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Tom [10]
3 years ago
10

Solving 2-D Motion

Physics
2 answers:
Lesechka [4]3 years ago
3 0

do you have a formulai can do this problem

zavuch27 [327]3 years ago
3 0

I don't claim to be an expert here, but if I am correct the Delta Y is .434

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a force of 25 newtons moves a box a distance of 4 meeters in 5 seconds.the work done on the box is ? NM and the power is. ? nm/
miskamm [114]

Answer:

The work done on the box is 100 Nm

The power is 20 Nm/s

Explanation:

There is a force 25 newtons moves a box a distance of 4 meters in

5 seconds

The work done on the box is the product of the force and the distance

that the box moves ⇒ <em>work = force × distance</em>

The force = 25 newtons

the distance = 4 meters

Work = 25 × 4 = 100 NM

<em>The work done on the box is 100 Nm</em>

<em></em>

The force moves the box 4 meters in 5 seconds

The power is the rate of work

<em>The power = work ÷ time</em>

The work = 100 Nm

The time = 5 seconds

The power = 100 ÷ 5 = 20 Nm/s

<em>The power is 20 Nm/s</em>

6 0
3 years ago
Both physical and chemical changes are accompanied by a change in the total energy of the system, which is divided into two broa
olga2289 [7]

Answer:

I believe <u>kinetic / potential</u>

Explanation:

8 0
2 years ago
A particle is moving with (SHM) of period 8.0s and amplitude5.0m
nadezda [96]

Answer:

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

Max speed = \frac{15\, \pi}{4} \,\, \frac{m}{s}

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

Explanation:

Given the description of period and amplitude, the SHM could be described by:

f(x)=5\,sin(\frac{\pi}{4}x)

and its angular velocity can be calculated doing the derivative:

f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)  and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = \frac{15\, \pi}{4} \,\, \frac{m}{s}

The acceleration is found from the derivative of the velocity expression, and therefore given by:

acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

6 0
3 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
3 years ago
Read 2 more answers
A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/
navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

\text{Initial Momentum} = \text{Final Momentum}

m_1u_1 +m_2u_2 = m_1v_1+m_2v_2

Here,

m_{1,2}= Mass of each object

u_{1,2}= Initial velocity of each object

v_{1,2}= Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

m_2u_2 = (m_1+m_2)v_f

Rearranging to find the final velocity

v_f = \frac{m_2u_2}{ (m_1+m_2)}

v_f = \frac{ 12500*7.8}{ 12500+7430}

v_f = 4.8921ft/s

The expression for the impulse received by the first car is

I = m_1 (v-u)

I = \frac{W}{g} (v-u)

Replacing,

I = \frac{12500}{32.2}(4.89-7.8)

I = -1129.65lb\cdot s

The negative sign show the opposite direction.

7 0
3 years ago
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