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viktelen [127]
3 years ago
12

A car accelerate uniformly from rest at 5m/s2 . Determine it's speed after 10s​

Physics
2 answers:
krek1111 [17]3 years ago
6 0

Answer:

50m/s.

Explanation:

Let's take acceleration as A and speed as S:

A = 5m/s²

S = A × 10s = 5 × 10 = 50m/s

The answer is 50m/s.

Sergeu [11.5K]3 years ago
4 0

Answer:

50m/s

Explanation:

Given:

a=5m/s^2

t=10s

Required:

v=?

Formula:

<em>a</em><em>=</em><em>v</em><em>/</em><em>t</em>

Solution:

a=v/t

v=a*t

v=5m/s^2*10s

v=50m/s

Hope this helps ;)❤❤❤

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a car takes 125 ft to brake from 60 to 0 mph. Assume that the acceleration of the Prius is constant while braking. Find how long
max2010maxim [7]

Answer:

2.84 seconds

Explanation:

t = ?

distance = 125

Velocity origianal = 60 m/hr = 88 ft/s

     AVERAGE velocity = 88/2 = 44 ft/s

44 t = 125

t = 125/44 = 2.84 s

7 0
2 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
3 years ago
Planet a and planet b are in circular orbits around a distant star. planet a is 7.8 times farther from the star than is planet
gogolik [260]
To find the ratio of planetary speeds Va/Vb we need the orbital velocity formula: 

V=√({G*M}/R), where G is the gravitational constant, M is the mass of the distant star and R is the distance of the planet from the star it is orbiting. 

So Va/Vb=[√( {G*M}/Ra) ] / [√( {G*M}/Rb) ], in our case Ra = 7.8*Rb 

Va/Vb=[ √( {G*M}/{7.8*Rb} ) ]  / [√( {G*M}/Rb )], we put everything under one square root by the rule: (√a) / (√b) = √(a/b) 

Va/Vb=√ [ { (G*M)/(7.8*Rb) } / { (G*M)/(Rb) } ], when we cancel out G, M and Rb we get:

Va/Vb=√(1/7.8)/(1/1)=√(1/7.8)=0.358 so the ratio of Va/Vb = 0.358.  
6 0
3 years ago
you toss a bowling ball straight up into the air with a speed of 2.1. how long does it take the bowling ball to reach its highes
omeli [17]
I am going to assume 2.1 metres per second and that we're rounding acceleration due to gravity to -10 metres per second squared. At the highest point, velocity is going to be 0. v= intial velocity + acceleration*time, sub in 0 for velocity, 2.1 for initial velocity and -10 for acceleration to get 0= 2.1-10t. Now solve for t. t=0.21 seconds.
3 0
3 years ago
Read 2 more answers
Help please! this is science :)
SSSSS [86.1K]

Answer: Answer is B

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