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grigory [225]
3 years ago
11

Ethan drove a 1500 kg car around a circular track with a radius of 75 m. If the magnitude of the centripetal acceleration is 2 m

/s2, what is the car's speed?
A)2.25 m/s
B)20 m/s
C)37.5 m/s
D)150 m/s ​
Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

12.25 m/s

Explanation:

The following data were obtained from the question:

Mass (m) = 1500 Kg

Radius (r) = 75 m

Centripetal acceleration (a) = 2 m/s²

Velocity (v) =?

Centripetal acceleration (a) is related to the velocity (v) and radius (r) according to the following formula:

a = v²/r

Thus, we can obtained the velocity of the car by using the above formula as illustrated below:

Radius (r) = 75 m

Centripetal acceleration (a) = 2 m/s²

Velocity (v) =?

a = v²/r

2 = v²/75

Cross multiply

v² = 2 × 75

v² = 150

Take the square root of both side.

v = √150

v = 12.25 m/s

Thus, the speed of the car is 12.25 m/s.

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What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving
lions [1.4K]

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

3 0
3 years ago
1 question, brainliest for whoever gets it right.
Kaylis [27]

Answer:

im pretty sure the answer is c please mark me brainliest

6 0
3 years ago
A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r
KIM [24]

Answer:

\omega_f = 1.08 rad/s

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

L_i = L_f

I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f

here we know that

I_o = \frac{1}{2}mR^2

I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2

m_1 = 9.41 kg

r_1 = \frac{4}{5}(1.61) = 1.29 m

m_2 = 21.1 kg

Now from above equation

118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f

118.1(1.73) = 188.45\omega_f

\omega_f = 1.08 rad/s

8 0
3 years ago
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the
goldfiish [28.3K]

Answer:

r = 0.405m = 40.5cm

Explanation:

In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:

F_c=ma_c=m\frac{v^2}{r}       (1)

Fc: centripetal acceleration (tension force on the string) = 12N

m: mass of the ball = 60g = 0.06kg

r: length of the string = ?

v: linear speed of the ball = 9.0m/s

You solve for r in the equation (1) and replace the values of the other parameters:

r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m

The length of the string between Wanda and the ball is 0.405m = 40.5cm

7 0
3 years ago
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