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3 years ago

Which process will be MOST AFFECTED by the heat?

2 answers:

6 0

Here, every process is natural which doesn't depend on temperature that much except condensation which means, conversion of vapor to liquid. so it would affected mostly by the heat

In short, Your Answer would be Option C

Hope this helps!

Alex Ar [27]3 years ago

5 0

The answer is precipitation

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A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0

3 years ago

What is the objects average velocity?

lyudmila [28]

Answer:

answer should be 10 because the line goes from (0,0) then to (1,10) and so on

5 0

3 years ago

A simple electric motor consists of a 220-turn coil, 4.4 cm in diameter, mounted between the poles of a magnet that produces a 9

Verdich [7]

Answer:

$M = 5.01\ A.m^2$

Explanation:

It is given that,

Number of turns in the coil, N = 220

Diameter of the coil, d = 4.4 cm

Radius of the coil, r = 2.2 cm = 0.022 m

Magnetic field produced by the poles of magnet, $B=96\ mT=96\times 10^{-3}\ T$

Current flowing in the coil, I = 15 A

Let M is the coil's magnetic dipole moment. Its formula is given by :

$M=N\times I\times A$

$M=220\times 15\times \pi (0.022)^2$

$M = 5.01\ A.m^2$

So, the coil's magnetic dipole moment is $5.01\ A.m^2$ . Hence, this is the required solution.

4 0

3 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
\ J$

$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

3 0

3 years ago

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

6 0

3 years ago