Choose the law that BEST explains the example:

Two forces P and Q act on an object of mass 15.0 kg with Q being the larger of the two forces. When both forces are directed to

Rom4ik [11]

Answer:

P=6.25N and Q=16.25N

Explanation:

In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).

Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:

$\Sigma F=ma$

We can assume there will be only the two mentioned forces P and Q, so

the sum of forces will be:

P+Q=ma

$P+Q=(15kg)(1.50m/s^{2})$

P+Q=22.5N

We can do the same analysis for the second free body diagram:

$\Sigma F=ma$

$Q-P=(15kg)(0.7m/s^{2})$

Q-P=10.5N

so now we have a system of equations we can solve by elimination:

Q+P=22.5N

Q-P=10.5N

Now, we can add the two equations together so the P force is eliminated, so we get:

2Q=32.5N

now we can solve for Q:

$Q=\frac{32.5N}{2}$

so

Q=16.25N

Now we can use any of the equations to find P.

Q+P=22.5N

P=22.5N-Q

when substituting for Q we get:

P=22.5N-16.25N

so

P=6.25N

5 0

4 years ago

A 0.366 kg metal cylinder is placed inside the top of a plastic tube, the lower end of which is sealed off by an adjustable plun

DochEvi [55]

Answer:

$a = 32.6 m/s^2$

Explanation:

As we know that pressure between the cylinder and plunger is increased by 1.59 times

So this will make a net force upwards on the cylinder which is given as

$F = \Delta P A$

now we will have

$\Delta P = P_2 - P_1$

Here initial pressure is given as

$P_1 = P_o + \frac{mg}{A}$

now new pressure is given as

$P_2 = 1.59 P_1$

so we have force on the cylinder given as

$F = P_2A - mg - P_oA$

$F = 1.59(P_0 + \frac{mg}{A})A - (mg + P_0A)$

$F = 0.59(1.01 \times 10^5 \times \pi(7.24 \times 10^{-3})^2 + 0.366(9.81))$

$F = 11.93 N$

now the acceleration is given as

$F = ma$

$11.93 = 0.366 a$

$a = 32.6 m/s^2$

4 0

3 years ago

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

Katarina [22]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

mass of skier, $m=70\ kg$

initial velocity of skier, $u=4\ m.s^{-1}$

height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

final velocity of skier before coming in contact of spring:

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

Now the time taken by the skier to reach down:

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

Now we calculate force using Newton's second law:

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

∴Compression in spring before the skier momentarily comes to rest:

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

4 0

3 years ago

If you used 650 W of power lifting a 300 N weight in 2 seconds how high did you lift?

dimulka [17.4K]

Answer:

4.33m

Explanation:

Power = work done/ time

work done = power × time =650 × 2 = 1300J

work done = force × distance

distance = work done/force

distance = 1300/300 = 4.33m

7 0

3 years ago

Kinetic and

beks73 [17]

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

5 0

3 years ago