3 years ago

What condition is necessary for an object to make a good reference point?

Physics

1 answer:

777dan777 [17]3 years ago

8 0

The object is fixed relative to the motion you are trying to describe.

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Which of the following is an example of aerobic activity?

mafiozo [28]

An example of an aerobic activity would be A, Cross-country skiing

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3 years ago

a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft

vlabodo [156]

Answer:

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3 years ago

.2, A car starting from rest has an acceleration of

riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

= 0 + 0.5 × 5

= 2.5 m/s

s = ut + 1/2 at²

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3 years ago

I started to solve this problem but I’m not sure on what to do next.

snow_tiger [21]

ANSWER:

3408.81 kg

STEP-BY-STEP EXPLANATION:

Given:

v = 111 m/s

Ek = 21000000 J

We have that the formula for kinetic energy is as follows:

$E_k=\frac{1}{2}\cdot m\cdot v^2$

We substitute the values given in the exercise and solve for m (mass)

$\begin{gathered} 21000000=\frac{1}{2}\cdot m\cdot111^2 \\ m=\frac{21000000\cdot2}{111^2} \\ m=3408.81\text{ kg} \end{gathered}$

The mass of the helicopter is 3408.81 kilograms.

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1 year ago

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

LUCKY_DIMON [66]

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power ,P = F .v

F=force

v=velocity

v= 100 km/h

$v=\dfrac{1000}{3600}\times 100\ m/s$

v=27.77 m/s

75 x 1000 = F x 27.77

$F= \dfrac{75000}{27.77}\ N$

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

3 0

3 years ago