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mezya [45]
3 years ago
14

Detlev walks 1000 meters north to the store and walks back 900 meters south to his friends house. The friends house is 100 meter

s to the north of Detlev’s house. What was Detlev’s Travel Distance?
Physics
1 answer:
lilavasa [31]3 years ago
3 0
1000 is the correct answer
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When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

P=\frac {W}{t}=\frac {FS}{t} where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328  W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

F=\frac {P}{V}

F=\frac {125328}{23}=5449.043478  N

F=5449 N

7 0
3 years ago
When light reaches the end of a barrier on the edge of an opening it diffracts, which means...
Mama L [17]
Hey there Evan!

Let's remember, these are (light waves). So, we you learn about waves, you would remember that when light does reach an end, they will most likely just (bounce back). That's all it would actually do.

I Hope this helps you!
3 0
3 years ago
If a system's internal energy increases by 250 kJ after the addition of375 kJ of energy as heat, what was the value ofthe work i
nata0808 [166]

Answer:

-125 kj

Explanation:

A system internal energy increases by 250kj after an additional 375kj

Therefore the value of the work in process can be calculated as follows

= 250kj-375kj

= - 125 kj

Hence the value of the work in process is - 125 kj

7 0
3 years ago
What is the motion of the particles in this kind of wave?
MatroZZZ [7]
I would say C but I’m not complete sore
8 0
3 years ago
Read 2 more answers
A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2
Masteriza [31]

Answer:

The answer is "512 J".

Explanation:

bullet mass m_1 = 10 g= 10^{-2} \ kg\\\\

initial speed u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\

block mass m_2 = 4\ Kg

initial speed v_2 =-4.2 \frac{m}{s}

final speed v_2= 0

Let v_1 will be the bullet speed after collision:

throughout the consevation the linear moemuntum

\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\

                = 320 \frac{m}{s}

The kinetic energy of the bullet in its emerges from the block

k=\frac{1}{2} m_1 v_1^2

   =\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J

3 0
3 years ago
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