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Flauer [41]
2 years ago
12

The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of t

he cart are 11.0 ft/s at point A and 18.0 ft/s at point C. The cart takes 5.00 s to go from point A to point C, and the cart takes 1.30 s to go from point B to point C. What is the cart's speed at point B
Physics
1 answer:
Tems11 [23]2 years ago
5 0

Answer:

The speed at B is 16.18 ft/s .

Explanation:

Speed at A, u = 11 ft/s

Speed at C, v' = 18 ft/s

Time from A to C = 5 s

Time from B to C = 1.3 s

Let the speed of car at B is v.

Let the acceleration is a.

From A to B

Use first equation of motion

v = u + a t

18 = 11 + a x 5

a = 1.4 ft/s^2

Let the time from A to B is t' .

t' = 5 - 1.3 = 3.7 s

Use first equation of motion from A to B

v = 11 + 1.4 x 3.7 = 16.18 ft/s  

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Find equivalent resistance. <br><br>Answer asap and please, please don't spam.​
Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega

So, the equivalent resistance s 4.77 ohms.

4 0
3 years ago
Which of the following statements is true about magnetic fields produced by current-carrying wires?
Anit [1.1K]
<span>The larger the current flowing in a wire, the stronger the magnetic field
is that surrounds the wire. 

That's why, if you want to make an electromagnet stronger, one way to
do it is to add another battery.  By increasing the voltage, you'll increase
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7 0
3 years ago
Read 2 more answers
A 600 kg car is at rest, and then accelerates to 5 m/s.
Crank

Answer:

0

7500J

7500J

Explanation:

Given parameters:

Mass of car  = 600kg

Velocity  = 5m/s

Unknown:

Original kinetic energy  = ?

Final kinetic energy  = ?

Work used  = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

                K.E  = \frac{1}{2}  m v²

  where m is mass

              v is velocity

original kinetic energy;

  The car started at rest and v = 0, therefore K.E  = 0

Final kinetic energy;

           K.E  = \frac{1}{2}  x 600 x 5²   = 7500J

Work done:

   Work done  = Final K.E  - Initial K.E  = 7500 - 0 = 7500J

   

3 0
2 years ago
When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup
Xelga [282]

Answer:

E = 169.34 J

Explanation:

First, we need to find the frictional force between the back tire and the road. For that purpose, we use the following formula:

f = μR = μW

f = μmg

where,

f = frictional force = ?

μ = coefficient of friction between tire and road = 0.8

g = 9.8 m/s²

m = mass supported by back tire = (0.5)(90 kg) = 45 kg

Therefore,

f = (0.8)(45 kg)(9.8 m/s²)

f = 352.8 N

Now, for the heat energy we use the formula of work. Because, thermal energy will be equal to work done by frictional force:

E = W = fd

where,

E = Thermal Energy = ?

f = frictional force = 352.8 N

d = displacement = 48 cm = 0.48 m

Therefore,

E = (352.8 N)(0.48 m)

<u>E = 169.34 J</u>

5 0
3 years ago
A particle is projected at an angle 60 degrees to the horizontal with a speed of 20m/s. (i) calculate total time of flight of th
Kisachek [45]

Answer:

Time of flight=3.5 seconds

Speed at maximum height is 0

Explanation:

Φ=60°

initial velocity=u=20m/s

Acceleration due to gravity=g=9.8 m/s^2

Total time of flight=T

Final speed=v

question 1:

T=(2 x u x sinΦ)/g

T=(2 x 20 x sin60)/9.8

T=(2 x 20 x 0.8660)/9.8

T=34.64/9.8

T=3.5 seconds

Question 2

Speed at maximum height is 0

8 0
3 years ago
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