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4 years ago

A train travels 58 km in two hours and 71 km in four hours what is its average speed

1 answer:

6 0

Speed is given as distance divided by time. I think that first you should simplify the two rates 58/2=29km/hr and 71/4=17.75km/hr After that find the mean or average which is (29+17.75)/2=23.375km/hr
Therefore it is safe to say that the average speed of the train is 23.375km/hr. I am not sure about this.

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Simon is riding a bike at 12 km/h away from his friend Keesha. He throws a ball at 5 km/h back to Keesha, who is standing still

pychu [463]

So there are different ways this could be solved. I'll do try to explain it the way I was taught...

Simon is riding his bike at 12 km/hr relative to the sidewalk, away from where Keesha is.

Simon throws the ball at Keesha, at 5 km/hr.

Keesha sees the ball approaching her at (12-5) = 7 km/hr relative to the ground to her.

Therefore the answer is: 7 km/hr

3 0

3 years ago

Does an object travel farther on a smooth or slippery surface or on a rough surface? Why?

Vitek1552 [10]

Answer:

There is much more friction on the rough surface than there is on the smooth surface.

Explanation:

8 0

3 years ago

How do you find the acceleration of an object?

artcher [175]

Rearrange the equation F = ma to solve for acceleration. You can change this formula around to solve for acceleration by dividing both sides by the mass, so: a = F/m. To find the acceleration, simply divide the force by the mass of the object being accelerated.

Hope i helped :)

4 0

3 years ago

Read 2 more answers

A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under

Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ = (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] = 1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ = 1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)] = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ = (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0

4 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$

The length of the lake is 499 m.

8 0

3 years ago