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Afina-wow [57]
3 years ago
7

On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per

hour, then in a few seconds, you speed up to 60 miles per hour and drive at this speed for 1 hour.
What was the x component of average velocity for the 6-hour period, using the fundamental definition of average velocity, which is the displacement divided by the time interval?
Physics
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

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A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra
ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

  H = \dfrac{u^2sin^2 \theta}{2g}

  H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}

         H =  0.889 m

the maximum height of the skateboard above the ground

         = 1 + 0.889

         = 1.889 m

b) time to reach the height

   t = \dfrac{u\ sin\theta}{g}

   t = \dfrac{5.1\ sin55^0}{9.8}

          t = 0.426 s

horizontal distance = u cos θ × t

                                = 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m

7 0
3 years ago
The amount of work done by two boys who apply 200 N of force in an
Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.  

           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0
3 years ago
A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start
Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4  m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction =  PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno. 

4 0
3 years ago
The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth),
SVETLANKA909090 [29]

Answer:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

V_e=\sqrt{\frac{2GM}{R}}

Escape velocity at the surface of earth

11=\sqrt{\frac{2GM}{R}}--------------------1

If Escape velocity is three times and the radius is also the three times

V_e_2=\sqrt{\frac{2G(3M)}{3R}}

V_e_2=\sqrt{\frac{2GM}{R}}=V_e

i.e. V_e_2=11 km/s

5 0
3 years ago
a snail takes 16 minutes 40 seconds to cover a distance of 1 m calculate the average speed of the swimmer​
matrenka [14]
The snail’s speed is 0.001042. Hope this helps!

8 0
2 years ago
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