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3 years ago

13

If the height of the ramp was 1.2 m above the floor, how long would it take for the marble to hit the ground after it left the r

amp?

1 answer:

Evgesh-ka [11]3 years ago

8 0

It would take at less 10 minte i guess this the right awnser

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In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pu

pashok25 [27]

Answer:

The answer is c) same.

Explanation:

When a force is applied at some point on a rigid body, the body makes a rotational movement about the axis. Said property of the force applied to rotate the body is calculated with a magnitude known as torque, which is defined as the moment of force at a capacity of that force to make a rotation around a specific point.

for case 1, the magnitude of the torque is equal to:

τ = F*(L/2)

For case 2, the torque magnitude equals:

τ = F*sin(30°) = F*(L/2)

It can be seen that in both cases it is the same torque.

7 0

2 years ago

Kindly answer the question about Work and Power. Image is attached below.

natita [175]

Answer:

89600 Watts.

Explanation:

From the question given above, the following data were obtained:

Mass (m) of speedboat = 2800 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Power (P) expended by the speedboat =?

Next, we shall determine the acceleration of the speedboat. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Acceleration (a) =?

v = u + at

16 = 0 + (a × 8)

16 = 0 + 8a

16 = 8a

Divide both side by 8

a = 16/8

a = 2 m/s²

The acceleration of the speedboat is 2 m/s²

Next, we shall determine the force exerted by the speedboat. This can be obtained as follow:

Mass (m) of speedboat = 2800 Kg

Acceleration (a) of speedboat = 2 m/s²

Force (F) of speedboat =?

F = ma

F = 2800 × 2

F = 5600 N

Finally, we shall determine the expended by the speedboat at maximum speed. This can be obtained as follow:

Force (F) of speedboat = 5600 N

Maximum velocity (v) = 16 m/s

Power (P) expended by the speedboat =?

P = F × v

P = 5600 × 16

P = 89600 Watts

Thus, the power expended by the speedboat is 89600 Watts.

4 0

3 years ago

A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet of mass 0.01 kg at a speed of 200 m/s. The recoil velocit

ozzi

Answer:

option D

Explanation:

given,

mass of the rifle,M = 2 Kg

mass of the bullet , m = 0.01 Kg

speed of the bullet, v = 200 m/s

recoil velocity ,v'= ?

initial velocity of bullet and rifle is zero.

using conservation of energy

Mu + m u' = M v'+ m v

0 + 0 = 2 x v' + 0.01 x 200

2 v' = -2

v' = -1 m/s

negative sign represent the velocity of rifle is in opposite direction of bullet.

hence, the correct answer is option D

5 0

3 years ago

5. Kelvin applied a rightward force of 302 N to a 29 kg desk to accelerate it across the floor. The

Naddik [55]

The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

$F_f = \mu mg$

where:

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration of gravity

In this problem we have

$\mu=0.750$

m = 29 kg

$g=9.8 m/s^2$

Therefore, the force of friction is

$F_f = (0.750)(29)(9.8)=213.2 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

3 0

2 years ago

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

7 0

3 years ago