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myrzilka [38]
3 years ago
14

If we approximate the rack to be completely flat and the racecar is travelling a constant 30.5 m/s around the turn, what forces

are responsible for the centripetal acceleration?
Physics
1 answer:
WITCHER [35]3 years ago
3 0

We have,

The race car is travelling a constant 30.5 m/s around the turn. When the car is travelling around a turn, the centripetal force acts on it. The centripetal force is balanced by the force of friction between the car and the surface. Hence, frictional force is responsible for the centripetal acceleration.

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
Three common elements that can reorient their electrons into magnetic domains and become magnetic are iron, nickel, and ________
ohaa [14]

Answer:

Cobalt

Explanation:

8 0
3 years ago
When three identical bulbs of 60W,200V rating are connected in series at a 200V supply the power drawn by them will be?
Law Incorporation [45]

Answer:

P = 180 [w]

Explanation:

To solve this problem we must use ohm's law, which is defined by the following formula.

V = I*R & P = V*I

where:

V = voltage = 200[volts]

I = current [amp]

R = resistance [ohm]

P = power [watts]

Since the bulbs are connected in series, the powers should be summed

P = 60 + 60 + 60

P = 180 [watts]

Now we can calculate the current

I = 180/200

I = 0.9[amp]

Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.

And the power is defined by P = V*I

we know that the voltage is equal to 200[V], therefore

P = 200*0.9

P = 180 [w]

8 0
3 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

6 0
3 years ago
A train runs from New Delhi to Hyderabad it covers first of 420 km in 7 hours and the next distance of 360 km in 6 hours​
Mkey [24]

Explanation:

the answer is 16.7 miles or 60kmph

5 0
3 years ago
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