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Simora [160]
3 years ago
6

While waiting for his Mom to come out of the hairdresser's, Sean accidentally puts the car in gear and it begins to roll forward

. How far would the vehicle travel if it moved at 34 m/s for 2.5 s?
Physics
1 answer:
kicyunya [14]3 years ago
7 0

The distance travelled by the car is 85 m

Explanation:

The car in this problem here is travelling by uniform motion (so, at constant velocity and constant speed). Therefore, the distance covered by the car after a time t is given by:

d=vt

where

d is the distance covered

v is the speed

t is the time

For the car in this problem we have

v = 34 m/s

t = 2.5 s

Substituting into the formula, we find

d=(34)(2.5)=85 m

So, the car has travelled for 85 m.

Learn more about speed here:

brainly.com/question/8893949

#LearnwithBrainly

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3 years ago
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A ray of light strikes a plane mirror at an angle of incidence 35°. If the mirror is rotated through 10°.
xxTIMURxx [149]

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1 35°

2 20°

3 50°

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7 0
3 years ago
Find the equivalent resistance.
frosja888 [35]

Answer:

18 Ω

Explanation:

As K and F are at the same voltage, we can redraw the diagram as in figure 2

Series resistances add directly, so we get figure 3

Adding parallel resistances gets us to figure 4

Now we can move two 6Ω resistances for clarification in figure 5

As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6

Add series resistances to get to figure 7

Add parallel resistances to get to figure 8

Add series resistances to get to figure 9

6 0
3 years ago
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A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another
Stella [2.4K]

Answer:

The chunk went as high as

2.32m above the valley floor

Explanation:

This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.

Applying the principle of energy conservation for the two ice we have based on the scenery

Momentum before impact = momentum after impact

M1U1+M2U2=(M1+M2)V

Given data

Mass of ice 1 M1= 5.20kg

Mass of ice 2 M2= 5.20kg

velocity of ice 1 before impact U1= 13.5 m/s

velocity of ice 2 before impact U2= 0m/s

Velocity of both ice after impact V=?

Inputting our data into the energy conservation formula to solve for V

5.2*13.5+5.2*0=(10.4)V

70.2+0=10.4V

V=70.2/10.4

V=6.75m/s

Therefore the common velocity of both ice is 6.75m/s

Now after impact the chunk slide up a hill to solve for the height it climbs

Let us use the equation of motion

v²=u²-2gh

The negative sign indicates that the chunk moved against gravity

And assuming g=9.81m/s

Initial velocity of the chunk u=0m/s

Substituting we have

6.75²= 0²-2*9.81*h

45.56=19.62h

h=45.56/19.62

h=2.32m

5 0
3 years ago
Find the work w1 done on the block by the force of magnitude f1 = 60.0 n as the block moves from xi = -3.00 cm to xf = 1.00 cm
Norma-Jean [14]
By definition, the work done by a force is given by:
 W1 = F1 * d

 Where,
 F: magnitude of force
 d: distance traveled.
 Substituting values we have:
 W1 = F1 * (xf - xi)

W1 = 60 * (0.01 - (-0.03))

 W1 = 60 * (0.01 + 0.03)

W1 = 60 * (0.04)

W1 = 2.40 J

 Answer: 
 the work w1 done on the block by the force of magnitude f1 = 60.0 n is:
 W1 = 2.40 J
6 0
3 years ago
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