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3 years ago

An object has a mass of 0.250 kg. What is the gravitational force of on the object by the earth?

Physics

1 answer:

sesenic [268]3 years ago

7 0

Answer:

2.4525 N

Explanation:

The earths gravity is 9.81 N/Kg

And so to work this out you would multiply 9.81 by 0.250 which equals to 2.4525N

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During the first 14 minutes of a 1.0 hour trip, a car has an average speed 36 km/h. What must the average speed in km/h of the c

oksian1 [2.3K]

Answer:

85.5 km/h

Explanation:

$t_{1}$ = time interval for first phase = 14 min = $\frac{14}{60}$ h = 0.233 h

$t_{2}$ = time interval for second phase = 46 min = $\frac{46}{60}$ h = 0.767 h

$v$ = average speed for the entire trip = 74 km/h

$v_{1}$ = average speed in first phase = 36 km/h

$v_{2}$ = average speed in second phase

$d_{1}$ = distance traveled in first phase

$d_{2}$ = distance traveled in first phase

average speed is given as

$v = \frac{d_{1} + d_{2}}{t_{1} + t_{2}}$

$v = \frac{v_{1} t_{1} + v_{2} t_{2}}{t_{1} + t_{2}}$

$74 = \frac{(36) (0.233) + v_{2} (0.767)}{0.233 + 0.767}$

$74 = (36) (0.233) + v_{2} (0.767)$

$v_{2} = 85.5$ km/h

4 0

3 years ago

Based on the table, what would you predict the speed of sound through ice would most likely be?

MatroZZZ [7]

Answer:

4000m/s

Explanation:

It would be this because sound travels faster through a solid rather than a liquid.

3 0

2 years ago

A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t

nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

$\epsilon = 1 - \dfrac{T_C}{T_H}$

$0.25 = 1 - \dfrac{T_C}{464}$

$T_C= (1 - 0.25) \times 464$

$T_C= 0.75 \times 464$

T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

$T_C= (1 - 0.42) \times 464$

$T_C= 0.58 \times 464$

$T_C= 269.12\ K$

8 0

3 years ago

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem

Shtirlitz [24]

Answer:

U = (ε0AV^2) / 2d

Explanation:

Where C= capacitance of the capacitor

ε0= permittivity of free space

A= cross sectional area of plates

d= distance between the plates

V= potential difference

First, the capacitance of a capacitor is obtained by:

C = ε0A/d.

Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor

Substitute in for C:

U = (ε0A/d) * V^2 / 2

Hence:

U = (ε0AV^2) / 2d

3 0

3 years ago

If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporizat

Zanzabum

Answer:

0.429 L of water

Explanation:

First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.

Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water

If this is true, then, we can assume that 1 kg of water = 1 L of water.

Knowing this, we have to use the expression to get energy which is:

Q = m * ΔH

Solving for m:

m = Q / ΔH

Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.

Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:

MM = 18 g/mol

The enthalpy in mass:

ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g

Finally, solving for m:

m = 970 / 2.261 = 429 g

Converting this into volume:

429 g = 429 mL

429 / 1000 = 0.429 L of water

3 0

3 years ago