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notsponge [240]
3 years ago
15

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st

art subscript, i, end subscript. Later, its speed increases to 4v_i4v i ​ 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Physics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

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Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters
amid [387]

Answer:

Part a)

t = 1.65 s

Part b)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

v = 27.3 m/s

direction of velocity is given as

[tex]\theta = 26.35 degree

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

a = \frac{3}{4}(9.81)

a = 7.36 m/s^2

now the vertical displacement covered by the canister is given as

y = 10 m

now by kinematics we have

y = \frac{1}{2}gt^2

10 = \frac{1}{2}(7.36)t^2

t = 1.65 s

Part b)

Horizontal speed of the canister is given as

v_x = 24.5 m/s

now the distance moved by it

x = v_x t

x = 24.5 (1.65)

x = 40.4 m

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

v_y = v_i + at

v_y = 0 + (7.36)(1.65)

v_y = 12.14 m/s

now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{24.5^2 + 12.14^2}

v = 27.3 m/s

direction of velocity is given as

\theta = tan^{-1}\frac{v_y}{v_x}

\theta = tan^{-1}\frac{12.14}{24.5}

[tex]\theta = 26.35 degree

6 0
3 years ago
What is work - energy theorem ??​
Elden [556K]

The work-energy theorem explains the idea that the net work - the total work done by all the forces combined - done on an object is equal to the change in the kinetic energy of the object. After the net force is removed (no more work is being done) the object's total energy is altered as a result of the work that was done.

This idea is expressed in the following equation:

is the total work done

is the change in kinetic energy

is the final kinetic energy

is the initial kinetic energy

mark me as brainliest ❤️

3 0
3 years ago
Read 2 more answers
A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels
katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

v = \sqrt{\frac{T}{\mu}}

Here,

v = Velocity

\mu= Linear density (Mass per  unit length)

T = Tension

Rearranging to find the Period we have that

T = v^2 \mu

T = v^2 (\frac{m}{L})

As we know that speed is equivalent to displacement in a unit of time, we will have to

T = (\frac{L}{t}) ^2(\frac{m}{L})

T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})

T = 5.54N

Therefore the tension is 5.54N

8 0
3 years ago
The normal formula to find force is F = m* a. What kind of math do you need to do
Oduvanchick [21]
To find the Mass of an object, you need to apply division.
Since Resultant Force = Mass X Acceleration

To find mass,
Mass = Force / Acceleration
3 0
3 years ago
How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied
adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0
3 years ago
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