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2 years ago

A car travels at 20km/h for 60 mins how far does it travel

Physics

1 answer:

r-ruslan [8.4K]2 years ago

8 0

Answer:

20km

Explanation:

20km/h means that for every 1h, the car traves 20km. 1h is 60 minutes so for every 60 minutes, the car travels 20km. Thus, it would travel 20km for 60 minutes.

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3 years ago

6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating

Anna007 [38]

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P = 55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

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Calculate output power

P = 55 kW

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= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

8 0

2 years ago

A single-turn current loop, carrying a current of 4.03 A, is in the shape of a right triangle with sides 68.1, 151, and 166 cm.

SashulF [63]

Answer:

Part a)

$F = 0$

Part b)

$F = 0.25 N$

Part c)

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

Explanation:

As we know that force on a current carrying wire is given as

$\vec F = i(\vec L \times \vec B)$

now we have

Part a)

current in side 166 cm and magnetic field is parallel

so we have

$F = i(\vec L \times \vec B)$

here we know that L and B is parallel to each other so

$F = 0$

Part b)

For 68.1 cm length wire we have

$F = iLB sin\theta$

here we know that

$cos\theta = \frac{68.1}{166}$

$\theta = 65.8$

so we have

$F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8$

$F = 0.25 N$

Part c)

For 151 cm length wire we have

$F = iLB sin\phi$

here we know that

$cos\phi = \frac{151}{166}$

$\theta = 24.5$

so we have

$F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5$

$F = 0.25 N$

Part d)

Net force on a closed loop in uniform magnetic field is always ZERO

$F_{net} = 0$

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2 years ago

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Gennadij [26K]

Answer:

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2 years ago

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