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Rashid [163]
3 years ago
13

A race car traveling at 10. meters per second accelerates at the rate of 1.5 meters per seconds while traveling a distance of 7,

467m. The final speed of the race car is approximately...?
1)1900m/s
2)150 m/s
3) 910 m/s
4) 44m/s
Physics
2 answers:
Arturiano [62]3 years ago
4 0

The final speed of the car is 2) 150 m/s

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 10 m/s

a=1.5 m/s^2

s = 7,467 m

Solving for v, we find the final velocity (and speed) of the car:

v=\sqrt{u^2+2as}=\sqrt{10^2+2(1.5)(7,467)}=150 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

aleksklad [387]3 years ago
4 0

Answer:44m/s

Explanation:

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A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic
Ilia_Sergeevich [38]

Answer:

\mu_k=0.27

Explanation:

According to the free body diagram, in this case, we have:

\sum F_x:-F_k=ma\\\sum F_y:N=mg

Recall that the force of friction is given by:

F_k=\mu_k N

Replacing and solving for the coefficient of kinetic friction:

-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}

Finally, we calculate \mu_k:

\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27

4 0
3 years ago
While spinning down from 500 rpm to rest, a flywheel does 3.9 kj of work. this flywheel is in the shape of a solid uniform disk
Ksivusya [100]

 

The answer is 4.0 kg since the flywheel comes to rest the kinetic energy of the wheel in motion is spent doing the work. Using the formula KE = (1/2) I w².

Given the following:

I =  the moment of inertia about the axis passing through the center of the wheel; w = angular velocity ; for the solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.

The radius = 1.2 m and KE = 3900 J

3900 J = (1/4) m (1.2)² (52.36)²

m = 3900 J / (0.25) (1.2)² (52.36)²

m = 3.95151 ≈ 4.00 kg

6 0
3 years ago
A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10
Alchen [17]

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

T = 39 days \times 24 hr \times 3600 s

T = 3369600 s

now the radius of the orbit is given as

r = 4.8 \times 10^8 m

so the total path length is given as

L = 2 \pi r

L = 2\pi (4.8 \times 10^8)

L = 3.015 \times 10^9

now the speed will be given as

v = \frac{L}{T}

v = \frac{3.015 \times 10^9}{3369600}

v = 895 m/s

7 0
3 years ago
¿Por qué es importante tu carrera en la sociedad? Menciona un mínimo de tres argumentos para sustentar tu respuesta y su aporte
agasfer [191]

Answer:

La carrera es importante para ganarse la vida, el respeto en la sociedad y cumplir el sueño.

Explanation:

La carrera es importante por las siguientes razones principales

a) Es una forma de ganar para alimentarse a sí mismo y a su familia y asegurar su futuro.

b) Permite ser considerado como una persona respetable en la sociedad.

c) Es importante ya que también brinda la oportunidad de cumplir el sueño.

3 0
3 years ago
Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.
Kamila [148]

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

8 0
3 years ago
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