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igomit [66]
3 years ago
7

C.

Physics
1 answer:
Anna [14]3 years ago
7 0

Answer:

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Một con lắc lò xo dao động điều hoà theo phương thẳng đứng, trong quá trình dao động của vật lò xo có chiều dài biến thiên từ 20
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6 0
3 years ago
11. A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds.
aleksklad [387]

Answer:

A) a = 7.89 m/s²

B) a = 3.51 m/s²

C) 4.38 m/s²

D) Frictional force

E) F_f = 5552.83 N

Explanation:

A) Formula for Force is;

F = ma

Where;

m is mass

a is acceleration

We are given;

F = 10,000 N

m = 1267 kg

Thus;

10000 = 1267a

a = 10000/1267

a = 7.89 m/s²

B) We are told the car covers 394.6 m in 15 seconds.

Using Newton's third equation equation of motion, we can find the actual acceleration.

s = ut + ½at²

u is zero since the object began from rest.

Thus;

S = ½at²

394.6 = ½ × a × 15²

a = 394.6 × 2/225

a = 3.51 m/s²

C) difference in accelerations = 7.89 - 3.51 = 4.38 m/s²

D) The force that caused the difference in acceleration is frictional force

E) To find the magnitude of the force that caused the difference in acceleration, we will use the formula;

F - F_f = ma

F_f = F - ma

Where F_f is the frictional force

Thus;

F_f = 10000 - 1267(3.51)

F_f = 5552.83 N

7 0
3 years ago
A ship sets sail from Rotterdam, The Netherlands, intending to head due north at 6.5 m/s relative to the water. However, the loc
sattari [20]

Answer:

Explanation:

velocity of ship with respect to water = 6.5 m/s due north

\overrightarrow{v}_{s,w}=6.5 \widehat{j}

velocity of water with respect to earth = 1.5 m/s at 40° north of east

\overrightarrow{v}_{w,e}=1.5\left ( Cos40\widehat{i} +Sin40\widehat{j}\right)

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth

\overrightarrow{v}_{s,w} = \overrightarrow{v}_{s,e} - \overrightarrow{v}_{w,e}

\overrightarrow{v}_{s,e} = 6.5 \widehat{j}- 1.5\left (Cos40\widehat{i} +Sin40\widehat{j}  \right )

\overrightarrow{v}_{s,e} = - 1.15 \widehat{i}+5.54\widehat{j}

The magnitude of the velocity of ship relative to earth is \sqrt{1.15^{2}+5.54^{2}} = 5.66 m/s

5 0
3 years ago
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