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2 years ago

15

If an object travels 245 km in 5 hours, what was the speed of the object?

1 answer:

seraphim [82]2 years ago

5 0

Answer:

49kmph

Explanation:

speed=distance/time = 245/5

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What s physics?

Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

7 0

3 years ago

Read 2 more answers

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

5 0

3 years ago

A force of 10N is required to stretch a spring from 20cm to 25cm. What is the spring constant in N/m2 Be careful of unit

kupik [55]

Answer:

C) 40 N/m

Explanation:

If we ASSUME that the spring is un-stretched at the zero cm position

k = F/Δx = 10/0.25 = 40 N/m

5 0

2 years ago

The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a

mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0

3 years ago

Read 2 more answers

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.60 m. How many times must he repeat this exercise to burn

Sholpan [36]

Answe

given,

mass of the bar, m = 30 Kg

distance of rise, h = 0.60 m

Assuming the efficiency = 25 %

energy from the pizza slice = 300 C = 1260 kJ

To consume Energy from the pizza bar is to be pulled several number of time.

( energy from pizza ) x (efficiency) = n m g h

n is the number of lift

( 1260 x 10³) x (0.25) = n x 30 x 9.8 x 0.6

$n = \dfrac{315\times 10^3}{176.4}$

n = 1786 times.

Weightlifter should lift bar 1786 times to burn off the energy.

8 0

3 years ago