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Diano4ka-milaya [45]
3 years ago
13

You drive your car in a straight line at 15 m/s for 10 kilometers, then at 25 m/s for another 10 kilometers.

Physics
1 answer:
Vikki [24]3 years ago
5 0

Answer:

A) Average speed = 18.75 m/s

B) More time is spent at 15 m/s than at 25 m/s.

Explanation:

Let the first distance be d1 and the second distance be d2.

We are given;

d1 = 10 km = 10000 m

d2 = 10 km = 10000 m

Speed; v1 = 15 m/s

Speed; v2 = 25 m/s

Now, the formula for distance is; Distance = speed x time

Thus:

d1 = v1 x t1

t1 = d1/v1 = 10000/15 = 666.67 seconds

Also,

d2 = v2 x t2

t2 = d2/v2 = 10000/25 = 400 seconds

Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s

From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;

More time at 15 m/s than at 25 m/s.

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IRINA_888 [86]

Answer:

Time : <u>7.96 s</u>

Distance Traveled : <u>357.8 m</u>  

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s    (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s    (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s    (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

<u>t₂ = 7.96 s</u>

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

<u>S = 357.8 m</u>

7 0
3 years ago
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net
kotykmax [81]

Answer:

0.29D

Explanation:

Given that

F = G M m / r2

F = GM(6m) / (D-r)2

G Mm/r2 = GM(6m) / (D-r)2

1/r2 = 6 / (D-r)2

r = D / (Ö6 + 1)

r = 0.29 D

See diagram in attached file

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How much kinetic energy the truck still has
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Answer:

47.5kJ

Explanation:

Before climbing the cliff

E_t = E_k

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At 2.

E_t = E_k+E_p

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timurjin [86]

Answer:

1.09527\times 10^{-8}\ N

Explanation:

q_1 = Mg ion = +2q

q_2 = F ion = -q

q = Charge of electron = 1.6\times 10^{-19}\ C

r = Distance between ions = 0.072+0.133\ nm

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electrical force is given by

F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N

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