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german
3 years ago
5

Two objects 5 kg and 7 kg are attached to the end of inextensible string which passes over frictionless pulley

Physics
1 answer:
Bumek [7]3 years ago
3 0

Answer:

ans:

tenson(T) = 20 N

acceleration (a) = 2.86 m/s

Explanation:

T + mg = Mg

T = Mg - mg

T = g( M - m )

T = 10× ( 7-5 )

T = 20 N

again;

T = 20

Ma = 20

a = 20 / 7

= 2.86 m/s

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A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
How far will a car travel in 30 min at 40 m/s? (kilometres)
kondor19780726 [428]

Answer:

72km

Explanation:

30 mins --> 30 x 60 s = 1800 s

Distance --> Speed x Time

= 40m/s x 1800s

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6 0
3 years ago
A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge
Kisachek [45]

Answer:

\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]

5 0
3 years ago
Which of the following statements regarding orbits is true?
Damm [24]

Answer:

the orbit of the a planet has an elliptical shape with the sun at the focus

Explanation:

3 0
3 years ago
A 420N force act on a 400N object, and the force is from the north.
belka [17]

Answer:

I believe its forward or south :D

Explanation:

7 0
3 years ago
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