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3 years ago

6

THE MASS AND WEIGHT OF A BOX IS 25 Kg & 245 N RESPECTIVELY WHEN ON EARTH. what is the value of mass and weight of the box wh

en it is brought to the moon ??? help this is due in like 5 freaking minutes im going to evaporate myself

1 answer:

Flauer [41]3 years ago

8 0

Answer:

33333

Explanati33on:33333333

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A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while

dem82 [27]

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

or

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$

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2 years ago

A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

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3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

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3 years ago

If a truck runs into a wall and stops, the truck loses momentum. Because

nignag [31]

The momentum goes to the wall

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2 years ago

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

klemol [59]

I think the answer for the question above its b 1.2

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3 years ago

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