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gregori [183]
3 years ago
11

How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s

Physics
1 answer:
Elis [28]3 years ago
7 0

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

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An electric hoist does 56,447 J of work in raising 115 kg load. How high (in meters) was the load lifted?
Ugo [173]

\\ \rm\rightarrowtail W=mgh

\\ \rm\rightarrowtail 56447=1150h

\\ \rm\rightarrowtail h=56447/11150

\\ \rm\rightarrowtail h=49.1m

8 0
2 years ago
If an object accelerates at 10 m/s2 for 4 seconds, how much will its velocity change by?
jek_recluse [69]

Answer:

40 m/s

Explanation:

Acceleration formula: a=\frac{\triangle v}{t}

Δv=at=10*4=40

5 0
2 years ago
A car from the beginning was traveling at 27.8m/s when he stepped on the brakes and stopped. According to the DMV, this would le
Yuliya22 [10]

Answer:

-6.44 m/s²

Explanation:

Given:

Δx = 60 m

v₀ = 27.8 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (27.8 m/s)² + 2a (60 m)

a = -6.44 m/s²

3 0
3 years ago
Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge
asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

E =\dfrac{kq}{r^2}

q =\dfrac{Er^2}{k}

inserting all the values

q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}

      q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0
3 years ago
Hey stob it.<br> Please help me.<br> Cmon help me.<br> Plz.
Anna [14]

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

6 0
2 years ago
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