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2 years ago

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Four friends push on the same block in different directions. Allie pushes on the block to the north with a force of 18 N. Bill p

ushes on the block to the east with a force of 14 N. Chris pushes on the block to south with a force of 23 N. Debra pushes on the block to the west with a force of 20 N. Assuming it does not move vertically, in which directions will the block move? north and west south and east south and west north and east

1 answer:

frozen [14]2 years ago

5 0

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.

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The process of changing a property of a wave to transmit information is called .

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Answer : The process of changing a property of a wave to transmit information is called Modulation.

Explanation :

Modulation is the process of changing the property of wave to transmit information. This is done with the help of modulator.

In modulation, the message signal is superimposed on a high frequency signal. A sine wave ( usually high frequency ) is used as a high frequency carrier wave.

Modulation can be done in many ways like :

(1) Frequency modulation

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What makes some collisions elastic and others inelastic?

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How much heat is released when 432 g of water cools down from 71'c to 18'c?

maria [59]

The heat released by the water when it cools down by a temperature difference $\Delta T$ is
$Q=mC_s \Delta T$
where
m=432 g is the mass of the water
$C_s = 4.18 J/g^{\circ}C$ is the specific heat capacity of water
$\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ}$ is the decrease of temperature of the water

Plugging the numbers into the equation, we find
$Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J$
and this is the amount of heat released by the water.

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2 years ago

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

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2 years ago

The electric force between two charged objects of charge +Q0 that are separated by a distance R0 is F0. The charge of one of the

vovikov84 [41]

Answer:

$F=3F_o(\frac{R_o}{R_{o2}})^2$

Explanation:

This problem is approached using Coulomb's law of electrostatic attraction which states that the force F of attraction or repulsion between two point charges, $Q_1$ and $Q_2$ is directly proportional to the product of the charges and inversely proportional to the square of their distance of separation R.

$F=\frac{kQ_1Q_2}{R^2}..................(1)$

where k is the electrostatic constant.

We can make k the subject of formula as follows;

$k=\frac{FR^2}{Q_1Q_2}...........(2)$

Since k is a constant, equation (2) implies that the ratio of the product of the of the force and the distance between two charges to the product of charges is a constant. Hence if we alter the charges or their distance of separation and take the same ratio as stated in equation(2) we will get the same result, which is k.

According to the problem, one of the two identical charges was altered from $Q_o$ to $3Q_o$ and their distance of separation from $R_o$ to $R_{o2}$ , this also made the force between them to change from $F_o$ to $F_{o2}$ . Therefore as stated by equation (2), we can write the following;

$\frac{F_oR_o^2}{Q_o*Q_o}=\frac{F_{o2}R_{o2}^2}{3Q_o*Q_o}.............(3)$

Therefore;

$\frac{F_oR_o^2}{Q_o^2}=\frac{F_{o2}R_{o2}^2}{3Q_o^2}.............(4)$

From equation (4) we now make the new force $F_{o2}$ the subject of formula as follows;

${F_oR_o^2}*{3Q_o^2}=F_{o2}R_{o2}^2*{Q_o^2}$

$Q_o$ then cancels out from both side of the equation, hence we obtain the following;

$3{F_oR_o^2}=F_{o2}R_{o2}^2.............(4)$

From equation (4) we can now write the following;

$F_{o2}=\frac{3F_oR_o^2}{R_{o2}^2}$

This could also be expressed as follows;

$F_{o2}=3F_o(\frac{R_o}{R_{o2}})^2$

3 0

2 years ago