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2 years ago

If a certain mass of mercury has a volume of 0.002m3 at a temperature of 20°C, what will be the volume at 50°C,

Physics

1 answer:

USPshnik [31]2 years ago

3 0

Set this up as a proportion.

.002 m^3/20 degrees = x/50 degrees

solve for x

x = .005 m^3

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18. Write conversion factors (as ratios) for the number of:

Romashka [77]

Answer:

Conversion tables show:

1 m = 1.09361 yds

1 Lit = .26418 gal = 1.05672 qt or 1 qt = .944632 Lit

1 lb = .45359 kg = 2.2046 Lbs / Kg

So X yds = X m * 1.09361 yds / m = 1.09361 * X yds

Likewise X Lit = X qt / 1.05672 qt/ Lit = X / 1.05672 Lit = .94632 X Lit

So X Lbs = X kg * 2.2046 Lbs / Kg = 2.2046 Lbs

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2 years ago

A rocket takes off from Earth's surface, accelerating straight up at 47.2 m/s2. Calculate the normal force (in N) acting on an a

lions [1.4K]

Answer:

Approximately $4.61\times 10^{3}\; {\rm N}$ upwards (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ .)

Explanation:

External forces on this astronaut:

Weight (gravitational attraction) from the earth (downwards,) and
Normal force from the floor (upwards.)

Let $(\text{normal force})$ denote the magnitude of the normal force on this astronaut from the floor. Since the direction of the normal force is opposite to the direction of the gravitational attraction, the magnitude of the net force on this astronaut would be:

$\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ .

Let $m$ denote the mass of this astronaut. The magnitude of the gravitational attraction on this astronaut would be $(\text{weight}) = m\, g$ .

Let $a$ denote the acceleration of this astronaut. The magnitude of the net force on this astronaut would be $(\text{net force}) = m\, a$ .

Rearrange $\begin{aligned}(\text{net force}) &= (\text{normal force}) - (\text{weight})\end{aligned}$ to obtain an expression for the magnitude of the normal force on this astronaut:

$\begin{aligned}(\text{normal force}) &= (\text{net force}) + (\text{weight}) \\ &= m\, a + m\, g \\ &= m\, (a + g) \\ &= 80.9\; {\rm kg} \times (47.2\; {\rm m\cdot s^{-2}} + 9.81\; {\rm m\cdot s^{-2}}) \\ &\approx 4.61 \times 10^{3}\; {\rm N}\end{aligned}$ .

3 0

1 year ago

You walk for a speed of 1.1 m/s (meters per second) for 6.0 seconds and

castortr0y [4]

Answer:

Average speed, As = 2.2 [m/s]

Explanation:

To solve these types of problems we must remember that the average of the speeds is determined by dividing the distance over time.

With the first speed and the time of 6 [s] we can calculate the distance.

V = x/t

where:

x = distance [m]

V = velocity = 1.1 [m/s]

t = time = 6 [s]

x1 = V*t

X1 = 1.1*6

X1 = 6.6 [m]

Now with the second velocity and 6 [s], we can calculate the second distance.

X2 = 3.3*6

X2 = 19.8 [m]

Now we have to calculate the average speed. The total distance is x = x1 +x2

X = 19.8 + 6.6 = 26.4 [m]

and the total time is 12 [s]

Therefore:

As = 26.4/12

As = 2.2 [m/s]

5 0

2 years ago

1. How much force would you have to apply to a 15kg object in order to accelerate it!<br> a 2 m/s?

mihalych1998 [28]

30N

Explanation:

Given parameters:

Mass of object = 15kg

Acceleration = 2m/s

Unknown:

Force = ?

Solution:

Force is given as the product of mass and acceleration:

F = m x a

m is the mass

a is the acceleration

Inputting the parameters:

F = 15 x 2 = 30N

The unit of force is newtons, N .

Learn more:

Force brainly.com/question/10470406

#learnwithBrainly

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2 years ago

What is the genotype for a woman with no freckles?

SIZIF [17.4K]

Well are having freckles recessive or dominant?

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2 years ago

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