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3 years ago

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

1 answer:

7 0

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

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How much force is needed to accelerate a 1750-kg car at a rate of 3 m/s2?

Rina8888 [55]

The force needed is 5250N

5 0

4 years ago

Please help me with this question

Ksju [112]

Answer:

i can not read that sorry

4 0

3 years ago

A car is moving at 10 m/s to the right. It accelerates for 10 s after which it is moving at 5 m/s to the left. What was the car'

NNADVOKAT [17]

Answer:

Acceleration, $a=-1.5\ m/s^2$

Explanation:

It is given that,

Initial velocity of the car, u = 10 m/s (in right)

Final velocity of the car, v = -5 m/s (in left)

Time taken, t = 10 s

Let a is the acceleration of the car. It can be calculated using the equation of kinematics. The equation is as :

$v=u+at$

$a=\dfrac{v-u}{t}$

$a=\dfrac{-5-10}{10}$

$a=-1.5\ m/s^2$

So, the acceleration of the car is $-1.5\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same

horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm= $8\times 10^{-2} m$

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid= $8\times 10^{-2} m$

Magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Using the formula

$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

Where $\mu_0=4\pi\times 10^{-7}$

$N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns$

6 0

3 years ago

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago