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3 years ago

Which force requires contact

Physics

2 answers:

guapka [62]3 years ago

6 0

Tension force? There ay be more than one, but that is all I can currently think of.

jekas [21]3 years ago

5 0

Gravity?????????? or maybe like any object

You might be interested in

" A bowl of soup at 200Á F. is placed in a room of constant temperature of 60Á F. The

Dahasolnce [82]

T(t)=60+140e−0.075t T(12)=60+140e−0.075∗12 T(12)=60+140e−0.9 T(12)=60+140(0.4065696597)
=116.84
So the temperature will be approximately 117 degrees

7 0

3 years ago

Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Leya [2.2K]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

final they both comes to rest so here we can say that final momentum must be zero

now we have

$34 v + 54 (3 m/s) = 0$

$v = -4.76 m/s$

6 0

3 years ago

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

Artyom0805 [142]

Answer:

$I=182.97\ kg-m^2$

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, $\tau=860\ N$

Angular acceleration of the shell, $\alpha =4.7\ m/s^2$

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

$\tau=I\alpha$

I is the rotational inertia of the shell

$I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2$

So, the rotational inertia of the shell is $182.97\ kg-m^2$ .

7 0

3 years ago

. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th

mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

$F \Delta t = m (v-u)$

where

F is the average force

$\Delta t$ is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

$\Delta t = 60.0 ms = 0.06 s$

Solving for F,

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N$

And since we are interested in the magnitude only,

F = 106.7 N

5 0

3 years ago

Read 2 more answers

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When the gas underg

GarryVolchara [31]

From the first law of thermodynamics, we use the equation expressed as:

ΔH = Q + W

where Q is the heat absorbed of the system and W is the work done.

We calculate as follows:

ΔH = Q + W
ΔH = 829 J + 690 J = 1519 J

Hope this answers the question. Have a nice day.

7 0

3 years ago

Read 2 more answers