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3 years ago

7

I apologise for asking too many questions. its because its a surprise test I didnt study. Thank you to the people who helped me.

love you.stau safe. this is the last question

1 answer:

Natali [406]3 years ago

7 0

Answer:

(a) gravity

(b) air resistance

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A 243 mL cup of whole milk contains about 45 mg of cholesterol. Express the cholesterol concentration of the milk in kilograms p

timama [110]

Answer: $0.1851 \frac{kg}{m^{3}}$

Explanation:

We have this concentration in units of $mg/ml$ :

$\frac{45 mg}{243 ml}$

And we need to express it in $kg/m^{3}$ , knowing:

$1 ml= 1 cm^{3}$

$1 m^{3}= (100 cm) ^{3}$

$1g = 1000 mg$

$1 kg = 1000 g$

Hence:

$\frac{45 mg}{243 cm^{3}} \frac{1 g}{1000 mg} \frac{1 kg}{1000 g} \frac{(100 cm) ^{3}}{1 m^{3}}=0.1851 \frac{kg}{m^{3}}$

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4 years ago

A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is

omeli [17]

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

$B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T$

4 0

3 years ago

Explain how star clusters help us understand the stages of stellar evolution

expeople1 [14]

Answer: More specifically, the term is typically defined as a period when gross domestic product (GDP) declines for two consecutive quarters. This prevailing line of thought was popularized by economist Julius Shiskin in 1974.

Explanation:

plz give me brainliest

5 0

3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/

Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt = mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0

3 years ago

Air at 20ºC with a convection heat transfer coefficient of 20 W/m2·K blows over a pond. The surface temperature of the pond is 2

dimulka [17.4K]

Answer:

The heat flux between the surface of the pond and the surrounding air is<em> 60 W/</em> $m^{2}$ <em> </em>

Explanation:

Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;

q = Q/A ...............................1

since we are working with the convectional heat transfer coefficient equation 1 become;

q = h ( $T_{sf} -T_{sd}$ ) ........................2

where q is the heat flux;

Q is the heat energy that will be transferred;

h is the convectional heat coefficient = 20 W/ $m^{2}$ .K;

$T_{sf}$ is the surface temperature = $23^{o}$ C 23°C + 273.15 = 296.15 K;

$T_{sd}$ is the surrounding temperature = $20^{o}$ C = 20°C + 273.15 = 293.15 K;

The values are substituted into equation 2;

q = 20 W/ $m^{2}$ .K ( 296.15 K - 293.15 K)

q = 20 W/ $m^{2}$ .K ( 3 K)

q = 60 W/ $m^{2}$

Therefore the heat flux between the surface of the pond and the surrounding air is 60 W/ $m^{2}$

3 0

4 years ago