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Rudiy27
3 years ago
15

A car speeds up from 12.0 m/s to 16.0 m/s in 8.0 s. the acceleration of the car is?

Physics
1 answer:
maria [59]3 years ago
6 0

Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2

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igor_vitrenko [27]
Inhales, digestive system
7 0
3 years ago
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A cyclist travels at 15 m/s during a sprint finish. What is this speed in km/h
g100num [7]

Answer:

54 km/hr

Explanation:

m/s to km/hr => 18/5

15 m/s to km/hr => 15 x 18/5 =>3 x 18 => 54km/hr

8 0
3 years ago
What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use
marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
3 years ago
How long does a 50 kW pump motor have to run to pump 500 cubic meters of water from a 180m deep mine
Ghella [55]

Answer:

t = 17658 s = 294.3 min = 4.9 h

Explanation:

The general formula for power is:

P = \frac{W}{t}

where,

P = Power of the Motor = 50 KW = 50000 W

W = Work Done by Motor = Change in P.E of Water = mgh

g = acceleration due to gravity = 9.81 m/s²

m = mass of  water = ρV

h = depth of water = 180 m

ρ = density of water = 1000 kg/m³

V = Volume of Water = 500 m³

t = time taken = ?

Therefore,

P = \frac{mgh}{t}=\frac{\rho Vgh}{t}\\\\t = \frac{\rho Vgh}{P}\\\\t = \frac{(1000\ kg/m^3)(500\ m^3)(9.81\ m/s^2)(180\ m)}{50000\ W}\\\\

<u>t = 17658 s = 294.3 min = 4.9 h</u>

6 0
3 years ago
In a lab, the mass of object A is 2.5 kg. Object A weighs:
dangina [55]

Answer:

25N

Explanation:

Assuming the lab is on earth:

w = mg = 2.5 (9.81) = 25N

8 0
3 years ago
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