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Shtirlitz [24]
3 years ago
12

if a car that is 100 feet in front of you on route 70 west slams on the brakes while you are traveling 65 miles per hour (95 fee

t per second) how much time do you have to hit the brakes and stop before rear ending them
Physics
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

t = 1.05 s

Explanation:

Given,

The distance between your vehicle and car, 100 ft

The constant speed of your vehicle, u = 95 ft/s

Since, the velocity is constant, a =0

If the car stopped suddenly, time left for you to hit the brake, t = ?

Using the second equation of motion,

                           S = ut + ½ at²

Substituting the given values in the equation

                           100 = 95 x t

                             t = 100/95

                               = 1.05 s

Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s

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Barry slides across an icy pond. The coefficient of kinetic friction between his
Soloha48 [4]

The force of friction acting on him is 122 N. The correct answer is option C

<h3>What is Friction ?</h3>

Friction is a force that opposes the motion of a static or a moving object.

Given that Barry slides across an icy pond. The coefficient of kinetic friction between his shoes and the ice is 0.15. If his mass is 83 kg

The given parameters are;

  • Mass m = 83 kg
  • Coefficient of kinetic friction μ = 0.15
  • Frictional force F_{r} = ?

The normal reaction N on the body = mg

N = 83 x 9.8

N = 813.4 N

The Frictional force formula is  F_{r} = μN

F_{r} = 0.15 x 813.4

F_{r} = 122.01 N

Therefore, the force of friction acting on him is 122 N approximately.

Learn more about Friction here: brainly.com/question/24338873

#SPJ1

4 0
2 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
Doss [256]
The first part of the question is 3,100 V.

The second part of the question is 200 V.
6 0
3 years ago
Read 2 more answers
A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t
Vanyuwa [196]

Answer:

 x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm  = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

  k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

W = \dfrac{1}{2}kx'^2

20= \dfrac{1}{2}\times 39.2\times x'^2

 x'² = 1.0204

 x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

 

7 0
3 years ago
A car increased its velocity to 62m/s in 10s starting from rest. Calculate the distance it covers during this time?
Roman55 [17]

Answer:

distance= velocity ×time

distance= 62×10

distance=620m

hope it helps you mate please mark me as brainliast

6 0
3 years ago
Read 2 more answers
A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr
quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = \frac{n}{2L} x \sqrt{\frac{T}{μ}}

f = \frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

8 0
3 years ago
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