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solong [7]
3 years ago
12

what is required to cause a body of mass 500g to accelerate uniformly from rest across a smooth horizontal surface so that it wi

ll cover a distance of 20 tometre in 4 seconds .​
Physics
1 answer:
Y_Kistochka [10]3 years ago
8 0

A constant force of 1.25N

Explanation:

This is a kinematics problem.

First find acceleration,

then use F = ma to find force.

Given:

mass = .5kg

delta x = 20m

t = 4s

a = ?

Friction = 0

From the kinematics equations:

delta x = Vi + (1/2)at^2

Plug in terms that are given:

20m = 0 + (1/2)a(4^2)

(2*20m)/(16s^2) = a

40m/(16s^2)= 2.5m/s^2

Now use F = ma to find force exerted on object.

F = (0.5kg)*(2.5m/s^2)

F = 1.25N

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A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
what would the mass be of an object that was moving at a velocity of 35 m/s and has a kinetic energy of 500j be?
bearhunter [10]

Answer:

0.816 kg

Explanation:

E_k=\frac{1}{2}mv^{2}\\

so m=\frac{2E_k}{v^2}=\frac{2\times500}{35^2}=0.816 kg

5 0
2 years ago
The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li
melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate Q of a hypodermic needle:

Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}  (1)

Where the dimensions of each one is:

Volume flow rate Q=\frac{L^{3}}{T}

Radius of the needle R=L

Length of the needle L=L

Pressures at opposite ends of the needle P_{2} and P_{1}=\frac{M}{LT^{2}}

Viscosity of the liquid \eta=\frac{M}{LT}

We need to find the value of n whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}  (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}  (3)

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}  (4)

As we can see n must be 4 if we want the exponent to be 3:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}  (5)

Finally:

\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}  (6)

8 0
3 years ago
(c) A moving train has a kinetic energy of 8.1 x 10(power of 6)J.
tamaranim1 [39]

the friction force provided by the brakes is 30000 N.

<h3>What is friction force?</h3>

Friction force is the force that opposes the motion between two bodies in contact.

To calculate the average friction force provided by the brakes, we apply the formula below.

Formula:

  • K.E = F'd............. Equation 1

Where:

  • K.E = Kinetic energy of the train
  • F' = Friction force provided by the brakes
  • d = distance

Make F' the subject of the equation

  • F' = K.E/d............ Equation 2

From the question,

Given:

  • K.E = 8.1×10⁶
  • d = 270 m

Substitute these values into equation 2

  • F' = (8.1 ×10⁶)/270
  • F' = 30000 N

Hence, the friction force provided by the brakes is 30000 N

Learn more about friction force here: brainly.com/question/13680415

8 0
2 years ago
Summarize: Based on what you have learned, how will the sound that the observer hears
Reil [10]

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

Explanation: Let assume that the observer is at a stationary position. The wavelength of the sound from the airplane reduces and the frequency increases as the plane is moving toward the observer. As the airplane passes by, that is, moving away from the observer, the frequency starts to reduce while the wavelength of the sound starts to increase.

The sound that the observer hears will change base on the illustration above.

3 0
2 years ago
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