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2 years ago

Father drove 176 km in 7 hours. For the first 92 km, he spent

Physics

1 answer:

Verizon [17]2 years ago

3 0

Answer:

28km/h

Explanation:

92km = 4 hours

176 - 92 = 84

84km = 3 hours

84/3 = 28

28km/h

Hopefully this helps you :)

pls mark brainlest ;)

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What is your displacement if you travel 5 miles north to the store

kumpel [21]

Answer:

5miles North

Explanation:

The displacement to the store is 5miles northward.

Displacement is the distance traveled in a specific direction. Displacement is a vector quantity.

A vector has both magnitude and direction.

The magnitude is the amount of that quantity

The direction is its orientation from a reference.

Therefore, the displacement is 5miles north

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2 years ago

The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th

ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, $B=2\ mT=2.2\times 10^{-3}\ T$

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

$B=\dfrac{\mu_oNI}{2\pi r}$

$r=\dfrac{\mu_oNI}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}$

r = 0.00523 m

$r=5.23\times 10^{-3}\ m$

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

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2 years ago

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The temperature inside my refrigerator is about 4 degrees C. If I place a balloon in my fridge that initially has a temperature

maksim [4K]

Convert the given temperatures from celsius to kelvin since we are dealing with gas.

To convert to kelvin, add 273.15 to the temperature in celsius.

T1 = 22 + 273.15 = 295.15 k

T2 = 4 + 273.15 = 277.15 k

V1 = 0.5 L

Let's find the final volume (V2).

To solve for V2 apply Charles Law formula below:

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1 year ago

I need help with number 6

Otrada [13]

I like your handwriting boy

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3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

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$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

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3 years ago

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