Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

$\sf\longrightarrow Weight = mg - ma \\$

$\sf\longrightarrow Weight = m ( g - a ) \\$

Mass of the man = 64.2 kg

$\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\$

$\sf\longrightarrow Weight = 64.2 * 6.44 N\\$

$\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\$

5 0

2 years ago

Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of

pantera1 [17]

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

$v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s$

(b) The angular velocity of disk 2 is :

$v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s$

(c) The moment of inertia for the two disk system is given by :

$I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2$

Hence, this is the required solution.

6 0

3 years ago

What automotive system keeps the vehicle's tires in contact with the road? A. The steering system B. The suspension sys

taurus [48]

The answer is D transmission

3 0

3 years ago

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