Ask question

Ask question

All categories

3 years ago

14

A body 'A' of mass 1.5kg travelling along the positive X-axis with speed of 4.5m/s collides with another body 'B' of mass 3.2kg,

which initially is at rest as a result of the collision, 'A' is deflected and moves in a speed of 2.1m/s in a direction which is at an angle of 30 degree below the X-axis. 'B' is set in motion at an angle Φ above the X-axis. calculate the velocity of 'B' after the collision.

1 answer:

xz_007 [3.2K]3 years ago

6 0

I already answered this question.
Please refer to this link brainly.com/question/8743596.

You might be interested in

At its natural resting length, a muscle is close to its optimallength for producing force. As the muscle contracts, the maximumf

lesantik [10]

Answer:

Figure E is the correct representation of the first part of the motion. When in a hanging position from the chin-up bar, the bicep muscles are stretched beyond their normal length already. So at this point they are at the peak of their capacity and you are at rest (this corresponds to the velocity v = 0 at t = 0). On contracting the bicep muscles and pulling your whole body up, you begin to gain speed and v increases. This increase in velocity is exponential. Soon the bicep muscles contract up to 80% their normal length reducing the force they can produce to keep you rising up to zero. The velocity change happens because the body is accelerating and the muscles can still supply a net force to lift you up. The acceleration is present because of this net force. The moment this force reduces to zero, the acceleration too reduces to zero. (From Newton's second law of motion). This reduction in acceleration is responsible for the reduction of the curvature of the v curve in figure E above. The point where the velocity becomes horizontal corresponds to the point where the muscles reach their maximum contraction unit and can supply no more net force and as a result no acceleration. This further results inba constant velocity which is the flat nature of the curve seen in diagram E.

Thank you for reading.

Explanation:

5 0

3 years ago

A mass is bobbing up and down on a spring. If you increase the amplitude of the motion, how does this affect the time for one os

NikAS [45]

Answer:

It will not change

Explanation:

The period of oscillation of a mass-spring system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

T is the period

m is the mass hanging on the spring

k is the spring constant

As we see from the formula, the period of oscillation does not depend on the amplitude of the motion: therefore, if we change the amplitude, the time for one oscillation will not change.

6 0

3 years ago

Calculating Welght<br> What is the weight of a 45 kg box?<br> N

Julli [10]

Answer:

Im not sure if this is right but here is the answer i think it is....So multiply 45 by 2.205 to get 99.225 pounds. Alternately, to convert from kg to Newtons, use the fact that 1 kg is 9.8 Newtons

Explanation:

7 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

lesantik [10]

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$ .

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$ .

The length of each side of the triangle, $l = 2.9~cm$ .

Consider a equilateral triangle $\bigtriangleup ABC$ , as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$ .

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for $q$ , $0.029~m$ for $l$ and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$

8 0

3 years ago

The ability of matter to easily combine chemically with other substances is know as?

skelet666 [1.2K]

Reactivity is the ability of matter that allows it to easily chemically combine with other substances.

4 0

3 years ago

Read 2 more answers