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Solnce55 [7]
3 years ago
7

The brakes of a car are applied, causing it to slow down at a rate of 10 ft/s2. Knowing that the car stops in 300 ft, determine

Physics
1 answer:
spayn [35]3 years ago
7 0

Answer:

(a) The velocity of the car before the brakes were applied is 77.46 ft/s

(b) The time required for the car to stop is 7.8 s

Explanation:

Given;

acceleration of the car, a = 10 ft/s²

distance traveled by the car, d = 300 ft

(a) the velocity of the car before the brakes were applied is given;

v² = u² + 2ad

v² = 0 + 2(10 x 300)

v² = 6000

v = √6000

v = 77.46 ft/s

(b) the time required for the car to stop

d = ut + ¹/₂at²

d = 0 +  ¹/₂at²

d = ¹/₂at²

t² = 2d / a

t = √ ( 2d / a)

t = √ ( 2 x 300 / 10)

t = 7.8 s

Therefore, the time required for the car to stop is 7.8 s

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A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s,
Andreyy89

Answer:

-450 m/s

Explanation:

Momentum is conserved.

p₀ = p

0 = (1.5 kg) (1.5 m/s) + (0.005 kg) v

v = -450 m/s

6 0
3 years ago
Define the difference between the rigid body problem and a single particle problem.
agasfer [191]
. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
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Can a body have constant speed and still be acclerating ? Give an Example
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Hi Pupil Here's Your answer :::





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3 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
A lamp hangs from the ceiling at a height of 2.6 m. The lamp has a mass of 3.8 kg. The screws holding the lamp break, and it fal
iVinArrow [24]

Answer:

Explanation:

Given height of lamp from the ceiling = 2.6m

mass of the lamp = 3.8kg

acceleration due to gravity = 9.81m/s²

As the body falls to the ground, it falls under the influence of gravity.

Gravitational potential energy = mass*acc due to gravity * height

Gravitational potential energy  = 3.8*2.6*9.81

Gravitational potential energy  = 96.923 Joules

b) Kinetic energy = 1/2 mv²

m = mass of the body (in kg)

v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion v^{2} = u^{2}+2gh

v^{2} = 0^{2}+2(9.81)(2.6) \\v^{2} = 51.012\\v =\sqrt{51.012}\\ v = 7.14m/s

Since mass = 3.8kg

K.E =  1/2 * 3.8 *7.14^{2}\\ K.E = 96.86Joules

c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s

4 0
3 years ago
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