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Zolol [24]
3 years ago
12

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

ve wheels is a constant 1300 N.What is the magnitude of the car's acceleration at t=13s?What is the direction of the car's acceleration at t=13s ? Give the direction as an angle from the r-axis.If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
8 0

Answer:

The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad

Explanation:

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2. A car accelerates uniformly from +10.0 m/s to +50.0 m/s over a distance of 225 m. How long did it take to go that distance? S
artcher [175]
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.

From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.

Hence P(t)=a*t^2/2+10t

Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T

Hence T=225/30=7.5

It took 7.5 seconds


7 0
3 years ago
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
joja [24]

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

6 0
3 years ago
Help!!! If anyone could do one of these, i'm confused on how I should write the equation down.
balu736 [363]

F=ma=m(change in velocity/change in time)

Number 1

F=ma

F=55kg(1.1ms^-1/1.6s)=37.8N

Number 2

F=ma

F=0.440kg(10ms^-1/0.02s)=220N

Number 3

F=ma

F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N

Any questions please feel free to ask.

4 0
3 years ago
The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg
nikitadnepr [17]
By applying Newton's second law of motion;

ma = mg - T

Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.

For the current case, the velocity is constant therefore,
a = 0

Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N

Tension in the cable is 1128.15 N.
8 0
3 years ago
A 1.0 kg block is attached to an unstretched
KonstantinChe [14]

Answer:

Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = 1 Nm.

Explanation:

Here, spring constant, k  = 50 N/m.

given block comes down eventually 0.2 m below.

here, g = 10 m/s.

let block be at a height h above the ground in figure 1.

⇒In figure 2, potential energy of the block-spring-Earth

system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.

⇒ Change in  potential energy of the block-spring-Earth

system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)

              =  (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.

4 0
3 years ago
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