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victus00 [196]
4 years ago
14

Tourists covered 255 km for a 4-hour ride by car and a 7-hour ride by train. what is the speed of the train, if it is 5 km/h gre

ater than the speed of the car?
Physics
1 answer:
LekaFEV [45]4 years ago
4 0
The distance covered by car is equal to (assuming it is moving by uniform motion) the product between the car's speed and the time of the car ride, 4 h:
S_c = v_c t_c
where
v_c is the car's speed
t_c = 4 h is the duration of the car ride

Similarly, the distance covered by train is equal to the product between the train's speed and the duration of the train ride, 7 h:
S_t = v_t t_t

The total distance covered is S=255 km, which is the sum of the distances covered by car and train:
S=255 km = S_c + S_t
which becomes
255 = 4 v_c + 7 v_t (1)
we also know that the train speed is 5 km/h greater than the car's speed:
v_t = 5 + v_c (2)

If we put (2) into (1), we find
255 = 4v_c + 7(5+v_c)
and if we solve it, we find
v_c = 20 km/h
v_t = 25 km/h

So, the car speed is 20 km/h and the train speed is 25 km/h.

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Answer:

(a).The maximum current in the wire is 4.217\times10^{5}\ A.

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(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

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Using formula of area

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Using formula of capacitor

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

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Using formula of resistivity

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}

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We need to calculate the voltage

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V=\dfrac{q}{C}

Put the value into the formula

V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}

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(a). We need to calculate the current

Using formula of current

I=\dfrac{V}{R}

I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}

I=421779.85\ A

I=4.217\times10^{5}\ A

(b). We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}

E=11.2\times10^{5}\ N/C

The electric field in the wire is 11.2\times10^{5}\ N/C.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

E=\dfrac{1}{2}CV^2

Put the value into the formula

E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2

E=1.126\times10^{-5}\ J

The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

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(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

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