The suspension system mounts the car's wheels solid on the frame. True or false?

Jin knows that the initial internal energy of a closed system is 78 J and the final internal energy is 180 J. He also knows that

lawyer [7]

As we know by the first law of thermodynamics

$Q = \Delta U + W$

here we know that

Q = heat given to the system

$\Delta U = U_f - U_i$

W = work done by the system

now here we can say

$\Delta U = 180 - 78 = 102 J$

$W = 64 J$

now we can say that heat will be given as

$Q = 64 + 102 = 166 J$

now here we can say that Jin does the error in his first step while calculation of change in internal energy as he had to subtract it while he added the two energy

So best describe Jin's Error is

B )For step 1, he should have subtracted 78 J from 180 J to find the change in internal energy.

8 0

3 years ago

In lightning , light is seen first and sound is heard later it is due to

Maksim231197 [3]

Answer:

option 4

Explanation:

Light's velocity in air ( 3 × 10^8 m/s ) is much greater than sound's velocity in air ( 343 m/s )

Hence due to difference in velocities , during lightning light is seen first & sound is heard later

8 0

3 years ago

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In the diagram, above, marker F is pointing to a __________, which are formed when meanders wear away at a narrow point and a po

castortr0y [4]

The answer is oxbow lake

4 0

3 years ago

Read 2 more answers

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the

ddd [48]

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

$F_{net} = 0$