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cestrela7 [59]
3 years ago
5

Wegener proposed the continental drift hypotheses suggesting that

Physics
1 answer:
kipiarov [429]3 years ago
3 0
There was a supercontinent called Pangea
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joshua started cycling at 5:15pm. by 8:09 pm, he has covered a distance of 7250m. what was joshua's average speed during that ti
ser-zykov [4K]

Answer:

2,500m/h

Explanation:

Average speed is calculated as total speed traveled divide by total time taken:

s_{avg}=\frac{d_{total}}{t_{total}}

Given total distance as 7250m and total time as 2009-1715hrs= 2hrs \ 54min=2.9hrs, speed is calculated as :

s_{avg}=\frac{d_{total}}{t_{total}}\\\\s_{avg}=\frac{7250m}{2.9hrs}\\\\=2500\ m/h

Hence, Joshua's speed is 2,500m/h

8 0
3 years ago
In a 5.00 km race, one runner runs at a steady 11.4 km/h and another runs at 14.7 km/h . How long does the faster runner have to
topjm [15]

Answer:

0.0986 h or 5 minutes 55 seconds.

Explanation:

Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.

Mathematically,

Speed = distance/time.

S = d/t ........................... Equation 1

making t  the subject  of the equation

t = d/S ......................... Equation 2

Form the question,

Time taken for the faster runner to reach the finish line

t₁ = d/S₁................... Equation 3

Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.

Given: d = 5.0 km, S₁ = 14.7 km/h.

Substituting into equation 3

t₁ = 5/14.7

t₁ = 0.340 h

Also,

t₂ = d/S₂................... Equation 4

Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.

Given: d = 5 km, S₂ = 11.4 km/h.

Substitute into equation 4,

t₂ = 5/11.4

t₂ = 0.4386 h.

The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340

The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.

8 0
4 years ago
1.A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upwar
Charra [1.4K]

Answer:

1. t = 3.27 seconds

2. y = 147.3 m

Explanation:

Newton's Laws of Motions.

y = v₁t + 1/2 at²

a = (v₂-v₁)/t

where

y = the vertical distance travelled

v₁ = the initial velocity

v₂ = the final velocity

t = the time

a = the acceleration

final velocity is equal to 0.

So, v₂ = 0.

a = (v₂-v₁)/t

a = (0-30)/t

a = -30/t

plugin values into the first equation:

y = v₁t + 1/2 at²

49 = 30t + 1/2 (-30/t)t²

49 = 30t -15t

49 = 15 t

t = 49/15

t = 3.27 seconds

2.

y = v₁t + 1/2 at²

a = -30/3.27

a = 9.2

y = 30(3.27) + 1/2(9.2) 3.27²

y = 147.3 m

6 0
3 years ago
A lumberjack is chopping wood with an ax. How does the ax make the lumberjack's job easier? A. The ax is heavy so it can split w
zhenek [66]

Answer: B

Explanation:

A machine multiplies the work output of a something by some amount. In this question, the ax is the machine. Thus, the ax is multiplying the work outputted by the lumberjack, so the answer is B. The reason it is not C is because there could be other forces interacting on the wood which the ax can affect.

5 0
3 years ago
Suppose that the period of a particular ideal mass-spring system is 5 s . What would be the period of the system if the mass wer
yan [13]

Answer: T2 = 7.07s

Explanation: The period of a loaded spring of spring constant k and mass m is given by

T= 2π √m/k

With 2π constant and k, it can be seen with little algebra that

T² is proportional to mass m

Hence (T1)²/m1 = (T2) ²/m2

Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.

By substituting, we have that

5²/m = (T2) ²/2m

25 / m = (T2) ²/2m

25 × 2m = (T2) ² × m

25 × 2 = (T2) ²

50 = (T2) ²

T2 = √50

T2 = 7.07s

5 0
4 years ago
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