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katen-ka-za [31]

2 years ago

13

Friction removes energy from objects in motion. Which statement best describes how this works?

1 answer:

xenn [34]2 years ago

8 0

<em>friction transforms KE into thermal energy (a)</em>

That's why, if it goes on long enough, the moving object actually gets warm.

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Contact force that acts to resist sliding between two touching surfaces

Black_prince [1.1K]

I think that the answer is friction

3 0

3 years ago

In the heat equation, what does c represent

Bingel [31]

Heat equation, Q = m.c.Δt
Here, c represents " the specific heat of the substance "

Hope this helps!

5 0

3 years ago

Read 2 more answers

A person throws a balloon straight down above level ground. The balloon bounces back up into the air. Drag force is significant

love history [14]

Explanation:

a) Balloon is being thrown down and is speeding up;

mg > $F_drag$

b) Balloon is in the air on its way down and moving with constant speed.

$F_drag$ =mg

c) The Balloon is on the ground and rest instantaneously

mg = Normal

d) Balloon is moving slowly downward;

e) Because, at the peak of trajectory drag force is 0.

Drag force = 0

5 0

3 years ago

You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and

andreev551 [17]

Answer:

v_avg = 37 km/h

Explanation:

To find the average velocity in the complete trajectory you use the following formula:

$v_{avg}=\frac{v_1+v_2}{2}$ ( 1 )

v1: velocity in the first part of the trajectory = 70 km/h

v2: velocity in the second part of the trajectory = ?

You calculate v2 by using the following equation for a motion with constant velocity:

$v_2=\frac{2.0km}{30min}*\frac{60min}{1h}=4\frac{km}{h}$

you replace the values of v1 and v2 in (1) and you obtain:

$v_{avg}=\frac{70km/h+4km/h}{2}=37\frac{km}{h}$

hence, the average velocity is 37 km/h

6 0

2 years ago

Read 2 more answers

Can someone please help me with these physics problems? I just don’t even know where to start.

KIM [24]

#1

for the block of mass 5 kg normal force is given as

$F_n = mg$

$F_n = 5*9.8 = 49 N$

friction force is given as

$F_f = \mu F_n$

$F_f = 0.1*49 = 4.9 N$

Net force is given as

$F_{net} = ma$

$F_{net} = 5*2 = 10 N$

now we know that

$F_{net} = F_{app} - F_f$

$10 = F_{app} - 4.9$

$F_{app} = 14.9 N$

#2

Normal force is given as

$F_n = mg$

$F_n = 6*9.8$

$F_n = 58.8 N$

now we know that

$F_{net} = F_{app} - F_f$

$F_{net} = 0$

as object moves with constant velocity

$F_{app} = F_f = 15 N$

now for coefficient of friction we can use

$F_f = \mu F_n$

$15 = \mu * 58.8$

$\mu = 0.255$

#3

net force upwards is given as

$F = 1.2 * 10^{-4} N$

mass is given as

$m = 7 * 10^{-5} kg$

now as per newton's law we can say

$F = ma$

$1.2 * 10^{-4} = 7 * 10^{-5} * a$

$a = 1.71 m/s^2$

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

$F_{net} = F_{app} - F_f$

$ma = F_{app} - F_f$

$5*6 = 40 - F_{f}$

$F_f = 40 - 30 = 10 N$

part b)

for coefficient of friction we can use

$F_f = \mu F_n$

$10 = \mu * F_n$

here normal force is given as

$F_n = mg = 5*9.8 = 49 N$

now we have

$\mu = \frac{10}{49} = 0.204$

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

$d = v_i*t + \frac{1}{2}at^2$

$20 = 0 + \frac{1}{2}a(5^2)$

$a = 1.6 m/s^2$

now net force is given as

$F_{net} = ma$

$F_{net} = 10*1.6 = 16 N$

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{0 - 25}{1.5} = -16.67 m/s^2$

now the force is gievn as

$F = ma$

$F = 5*16.67 = 83.3 N$

3 0

3 years ago