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denis-greek [22]
3 years ago
8

A crane moves a 250 kg scoreboard from the ground to the height of 100 m. What is the work done on the scoreboard?

Physics
2 answers:
Mazyrski [523]3 years ago
7 0
If you’re doing potential and kinetic energy then the answer is potential.
Sveta_85 [38]3 years ago
5 0

Answer:

You will have to do the last PE - the initial PE

Explanation:

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Two identical objects, A and B, are sitting on a table. If the net force on object A is 5 N and the net force on object B is 10
sveta [45]

If the net force on object A is 5 N and the net force on object B is 10 N, then object B will accelerate more quickly than object A provided the mass of both objects are same.

Answer: Option C

<u>Explanation: </u>

According to Newton’s second law of motion, any external force applied on an object is directly proportional to the mass and acceleration of the object. In order to state this law in terms of acceleration, it is stated that acceleration exhibited by any object is directly proportional to the net force applied on the object and inversely proportional to the mass of the object as shown below:

                      \text {Acceleration of the object } \propto \frac{\text {Net force on the object}}{\text {Mass of the object}}

So if two objects A and B are identical which means they have same mass, then the acceleration attained by the object will be directly proportionate to the net forces exerted on the objects only.

Thus if the force applied is more for one object, then the object will be exhibiting more acceleration compared to the other one. So as object B is experiencing a net force of 10 N which is greater than the net force experiences by object A, then the object B will be accelerating more quickly compared to the object A's acceleration.

7 0
3 years ago
An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine
liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, \theta = 30^{\circ}

Coefficient of kinetic friction, \mu_{k} = 0.100

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T        (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)

Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma

mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a

a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}

a = 10.45\ m/s^{2}

Now, the speed is given by the third eqn of motion with initial velocity being zero:

v^{2} = u^{2} + 2ad

where

u = initial velocity = 0

Thus

v = \sqrt{2ad}

v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s

3 0
3 years ago
A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is
ivanzaharov [21]
<span>anwser will be 

F = ma

where 

F = force exerted on the bullet 
m = mass of the bullet = 5 gm (given) = 0.005 kg. 
a = acceleration of the bullet 

Substituting appropriately, 

F = 0.005a --- call this Equation 1 

Next working equation is 

Vf^2 - Vo^2 = 2as 

where 

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given) 
Vo = initial velocity of bullet = 0 
a = acceleration of bullet 
s = length of the rifle's barrel 

Substituting appropriately, 

326^2 - 0 = 2(a)(0.83) 

a = 64,022 m/sec^2 

the anwser will be
Substituting this into Equation 1, 

F = 0.005(64,022) 

F =320.11 Newtons 

Hope this helps. </span><span>
</span>
8 0
3 years ago
A propagating wave in space with electric and magnetic components. These components oscillate at right angles to each other. It
zimovet [89]

Answer:

electromagnetic

Explanation:

8 0
4 years ago
Read 2 more answers
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
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