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stellarik [79]
2 years ago
6

Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 1.464 m/s/s. The acceleration

lasts for 10.8 s. What is the final speed of the car and how much ground does it cover during this acceleration?​
Physics
1 answer:
tamaranim1 [39]2 years ago
6 0

Answer:

A force 1500 n to a 100 kg car . what is the acceleration of the car

A sports car of mass 1000 kg can accelerate from rest to 27 m/s in 7.0 s. what is the average forwa...

A tow truck exerts a force of 2000 N on a car acceleration it at 1 m/s/s what is the mass of the ca...

A car starts from rest and accelerate at a constant rate. After the car has gone 50m it has a speed...

A force of 50 n is exerted to the right on a 15 kg car at the same time a 30 n force is exerted to ...

A 1100-kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat, ...

Your own car has a mass of 2000 kg if your car produces a force of 5000 n , how fast will it accele...

If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

A train consists of a caboose (mass = 1000 kg), a car (mass 2000 kg), and an engine car (mass 2000 ...

A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 12...

A car initially moving at 9 m/s begins accelerating at its maximum acceleration of 4 m/s2. if the c...

Car A runs a red light and broadsides Car B, which is stopped and waiting to make a left turn. Car ...

Speed Racer has car that accelerates at 5 m/s2 . If the car has a mass of 1000 kg, how much force d...

You push forward on a car with a force of 150 N, but the car does not move. What is the net force o...

How much time does it take for a car to accelerate from a standing start to 22.2 m/s if acceleratio...

A car has a mass of 1 500 kg and acceleration of 2.0 m/s2 what is the nat force acting on the car

A car starts from rest and travels for 5.0s with a uniform acceleration of 1.5 m/s^2. what is the f...

In a performance test, each of two cars takes 8.4 s to accelerate from rest to 27 m/s. car a has a ...

If a net force of 5.0 N forward acts on a 0.60 kg toy car what is its acceleration

Explanation:

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20 quantities and classified them in vector and scaler quantities​
ss7ja [257]

Answer:

A scalar quantity is defined as the physical quantity that has only magnitude, for example, mass and electric charge. On the other hand, a vector quantity is defined as the physical quantity that has both magnitude as well as direction like force and weight.

5 0
2 years ago
A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i
Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2    ;  μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0
3 years ago
Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the
Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
A bullet whose mass is 30.2 g leaves a rifle with a muzzle velocity of 1,000 m/s. It strikes a block of wood (mass 5 kg), initia
AveGali [126]
This problem here is an example of inelastic collision where kinetic energy is not conserved but momentum is. We calculate as follows:

m1v1 + m2v2 = (m1 + m2)v3
v3 = m1v1 + m2v2  / m1 + m2
v3 = (30.2)(1000) + (5000)(0) / (30.2 + 5000)
v3 = 6.00 m/s
5 0
3 years ago
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