Is it ever possible for the work done by friction to increase the kinetic energy of an object?

A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when

lana66690 [7]

Answer:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

3 0

2 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)

Soloha48 [4]

Answer:

Newton's second law: F=ma.

Force = mass x acceleration

This basically means, that the more massive something is, and the faster it is going, the more force the object will have.

So, the factors that can be used to find acceleration are:

F=ma

a=F/m

So, we see that the correct answer is A, mass and force.

Let me know if this helps!

8 0

2 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

4 0

3 years ago

The electric field is created by a voltage of 100 V. What's the

Ganezh [65]

Length of wire be l=1

Electric field intensity

$\\ \tt\hookrightarrow \dfrac{V}{l}$

$\\ \tt\hookrightarrow \dfrac{100}{l}$

$\\ \tt\hookrightarrow 100N/C$

3 0

2 years ago

A positive test charge q is released from rest at distance r away from a charge of +Q and a distance 2r away from a charge of +2

Svetradugi [14.3K]

Answer:

B. to the right

Explanation:

Given:

test charge +q

distance of the test charge from +Q, r

distance of test charge from +2Q, 2r

<u>Force on the test charge due to +Q:</u>

$F_1=k.\frac{Q.q}{r^2}$

<u>Force on the test charge due to +Q:</u>

$F_2=k.\frac{2Q.q}{(2r)^2}$

$F_2=k.\frac{Q.q}{2r^2}$

Since all the charges are positive here, so they will try to repel the test charge away. And the force due to charge +Q will be greater so initially the test charge will move rightwards away from the +Q charge.

3 0

3 years ago

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