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oee [108]
3 years ago
12

Determine the centripetal force on a vehicle rounding a circular curve with a radius of 80 m at a constant speed of 90 km/h if t

he vehicle's mass is 2,000 kg. (Hint: First express the velocity in meters per second)
Physics
1 answer:
quester [9]3 years ago
3 0
Mass of the vehicle = 2000 kg
Velocity of the vehicle = 90 km/hr
                                   = 25 m/s
Radius of the curve = 80 m
Then
Centripetal force = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 15625 kg m/s
                            = 15625 Newton second
I hope that this is the answer that you were looking for and the answer has come to your desired help. 
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A 50 kg child runs off a dock at 2.0 ms (horizontally) and lands in a waiting rowboat of mass 150 kg. At what speed does the row
Alex73 [517]

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

5 0
3 years ago
A stone is dropped from the upper observation deck of a tower, "500" m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dis
Yuri [45]

Answer:

h₍₁₎ = 495,1 meters

h₍₂₎ = 480,4 m

h₍₃₎ = 455,9 m

...

..

Explanation:

The exercise is "free fall". t = \sqrt{\frac{2h}{g} }

Solving with this formula you find the time it takes for the stone to reach the ground (T) = 102,04 s

The heights (h) according to his time (t) are found according to the formula:

h(t) = 500 - 1/2 * g * t²

Remplacing "t" with the desired time.

4 0
3 years ago
A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne
SVETLANKA909090 [29]

Given Information:  

Resistance of circular loop = R = 0.235 Ω 

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:  

Magnetic field = B = ?  

Answer:  

Magnetic field = 0.00145 T

Explanation:  

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space  and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

8 0
3 years ago
Find the force necessary to pull a 6 kg object 3 m/s2
mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

We have the final answer as

<h3>18 N</h3>

Hope this helps you

6 0
2 years ago
At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m
joja [24]

1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

x_c(t) = \frac{1}{2}at^2

where

a=5 m/s^2 is the acceleration

t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

x_t(t) = vt

where

v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the  same, so when

x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

x_c(8) = \frac{1}{2}(5)(8)^2=160 m

So, the car overtakes the truck 160 m far from the intersection.

2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

For the car in this problem, we have:

u = 0 (it starts  from rest)

a=5 m/s^2

And we know that the car overtakes the truck when

t = 8 s

Substituting into the equation,

v=0+(5)(8)=40 m/s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
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