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Suppose a door is 1 meter wide. Calculate the difference in the amount torque exerted on the door when someone pushes with a con

Yanka [14]

Answer:

Δτ = 50 N.m

Explanation:

The torque applied on an object is given by the product of the force applied on it and the perpendicular distance between the force and the axis of rotation of the object. That is:

τ = F r

where,

τ = Torque applied on the object

F = Force applied on it

r = distance from axis of rotation

FOR HANDLE SIDE OF DOOR:

τ₁ = F r₁

where,

τ₁ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₁ = distance from axis of rotation = 1 m

Therefore,

τ₁ = (100 N)(1 m)

τ₁ = 100 N.m

FOR MIDDLE OF DOOR:

τ₂ = F r₂

where,

τ₂ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₂ = distance from axis of rotation = 1 m/2 = 0.5 m

Therefore,

τ₂ = (100 N)(0.5 m)

τ₂ = 50 N.m

Now, the difference between the amount of torque in both cases is:

Δτ = τ₁ - τ₂

Δτ = 100 N.m - 50 N.m

Δτ = 50 N.m

6 0

3 years ago

If R = 12 cm, M = 520 g, and m = 20 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solv

timofeeve [1]

Answer:

v = 0.84 m/s

Explanation:

given,

R = 12 cm

M (mass of pulley )= 520 g

m (mass of block)= 20 g

s = 50 cm = 0.5 m

using conservation of energy

Potential energy = Kinetic energy

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

$I_{disk}= \dfrac{1}{2}MR^2$ and v = r ω

$m g h = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{1}{2}MR^2)(\dfrac{v}{R})^2$

$m g h = \dfrac{1}{2}mv^2 +\dfrac{1}{4}Mv^2$

$m g h = \dfrac{1}{2}v^2(m +\dfrac{1}{2}M)$

$v=\sqrt{\dfrac{2mgh}{m + 0.5 M}}$

$v=\sqrt{\dfrac{2\times 0.020 \times 9.8 \times 0.5}{0.02 + 0.5\times 0.52}}$

v = √0.7

v = 0.84 m/s

5 0

3 years ago

Consider a string of total length L, made up of three segments of equal length. The mass per unit length of the first segment is

zzz [600]

Answer:

Explanation:

Length of each segment is $\frac{L}{3}$

Speed of wave in first segment is $v_1=\sqrt{\frac{T_s}{\mu}}$

Speed of wave in second segment is $v_2=\sqrt{\frac{T_s}{2\mu}}$

Speed of wave in third segment is $v_3=\sqrt{\frac{T_s}{\frac{\mu}{4}}}=\sqrt{\frac{4T_s}{\mu}}$

Now time for the transverse wave to propagate is

$t=t_1 + t_2 + t_3\\=\frac{(\frac{L}{3})}{v_1}+\frac{(\frac{L}{3})}{v_2} + \frac{(\frac{L}{3})}{v_3}\\\\=(\frac{L}{3})(\frac{1}{\sqrt{\frac{T_s}{\mu}}} + \frac{1}{\sqrt{\frac{T_s}{2\mu}}} + \frac{1}{\sqrt{\frac{4T_s}{\mu}}})$

simplifying we get

$t=(\frac{3+2\sqrt{2}}{6})L\sqrt{\frac{\mu}{T_s}}$

3 0

3 years ago

A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on

Alisiya [41]

Answer:

Work required is 328387.5 Joules.

Explanation:

Given the following data;

Mass = 2085kg

Initial velocity, Vi = 13m/s

Final velocity, Vf =22m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

Where;

W is the work required.

Kf is the final kinetic energy possessed by the object.

Ki is the initial kinetic energy possessed by the object.

But Kinetic energy = ½MV²

W = ½MVf² - ½MVi²

Substituting into the equation, we have;

W = ½(2085)*22² - ½(2085)*13²

Simplifying the equation, we have;

W = 1042.5 * 484 - 1042.5 * 169

W = 504570 - 176182.5

W = 328387.5J

Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.

8 0

4 years ago

A current-carrying wire passes through a region of space that has a uniform magnetic field of 0.92 T. If the wire has a length o

alex41 [277]

Answer:

Current, I = 2.45 T

Explanation:

It is given that,

Magnetic field, B = 0.92 T

Length of wire, l = 2.6 m

Mass, m = 0.6 kg

We need to find the minimum current needed to levitate the wire. It is given by balancing its weight to the magnetic force i.e.

$Ilb=mg$

$I=\dfrac{mg}{lB}$

$I=\dfrac{0.6\ kg\times 9.8\ m/s^2}{2.6\ m\times 0.92\ T}$

I = 2.45 A

So, the minimum current to levitate the wire is 2.45 T. Hence, this is the required solution.

7 0

4 years ago