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3 years ago

How much force is needed to accelerate a 1000Kg car at a rate of 3m/s2?

Physics

2 answers:

11111nata11111 [884]3 years ago

8 0

F=force
M=mass
A=acceleration
F= M×A
F=1000×3
f=3000 kgm/s2
force required to accelerate car .

ikadub [295]3 years ago

6 0

According to Newton (2nd law), Force = (mass) x (acceleration)

Substitute what we know : Force = (1,000 kg) x (3 m/s²)

Do the arithmetic: Force = 3,000 kg-m/s² = 3,000 newtons

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The eiffel tower has a mass of 7.3 million kilograms and a height of 324 meters. its base is square with a side length of 125 me

uranmaximum [27]

Since the tower base is square with a side length of 125 m,

Therefore,

$(125\ m)^2+ (125\ m)^2=31250 m^2$

Square root of 31250 = 176.776953 (Diameter) , so this is the diameter of the cylinder to enclose it, and radius, r = 88.38834765 m and height, h = 324 m.

The volume of cylinder,

$=\pi r^2h=3.14(88.38834765 m)^2\times 324 m =7948168.803\ m^3$

Thus, the mass of the air in the cylinder,

$=1.225\ kg/m^3 \times 7948168.803\ m^3=9736506.78\ kg$

Hence, the mass of the air in the cylinder is this more than the mass of the tower.

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3 years ago

A 1400-kg car traveling east at 25m/s collides with a 1800-kg car traveling at a speed of 20m/s in a direction that makes angle

julia-pushkina [17]

M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
14*25 + 18*20[220o] = 14V + 18V.
350 + 360[220o] = 32V.
350 - 276-231i = 32V.
74 - 231i = 32V.
242.6[-72.2o] = 32V.
V = 7.6m/s[-72.2o]=7.6m/s[72o] S. of E.

7 0

3 years ago

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Zina [86]

Answer:

Explanation:

the first one is D

8 0

3 years ago

Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a di

Vlad1618 [11]

Answer:

4.44 kN in the opposite direction of acceleration.

Explanation:

Given that, the initial speed of the car is, $u=20m/s$

And the mass of the car is, $m=1000 kg$

The total distance covered by the car before stop, $s=45m$

And the final speed of the car is, $u=0m/s$

Now initial kinetic energy is,

$KE_{i}=\frac{1}{2}mu^{2}$

Substitute the value of u and m in the above equation, we get

$KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J$

Now final kinetic energy is,

$KE_{f}=\frac{1}{2}mv^{2}$

Substitute the value of v and m in the above equation, we get

$KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J$

Now applying work energy theorem.

Work done= change in kinetic energy

Therefore,

$F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN$

Here, the force is negative because the force and acceleration in the opposite direction.

6 0

3 years ago

A boat can travel in still water at 56 m/s. If the boats sails directly across a river that flows at 126 m/s. What is the boats

MAVERICK [17]

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

V² = 56² + 126²

V² = 19012

V = 137.9 m/s

7 0

3 years ago