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pishuonlain [190]
3 years ago
8

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

Physics
1 answer:
mihalych1998 [28]3 years ago
3 0

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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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
What is the relationship between the horizontal and vertical components of velocity for a projectile launched at an angle betwee
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Each is independent of the other -Each is dependent on each other -The horizontal component is dependent on gravitational acceleration -The horizontal component decreases with an increase in the vertical component

6 0
3 years ago
A 70 kg base runner begins his slide into second base while moving at a speed of 4.35 m/s. He slides so that his speed is zero j
sladkih [1.3K]

Answer

given,

Mass of the runner, M = 70 Kg

speed of the runner on the second base = 4.35 m/s

speed at the base = 0 m/s

Acceleration due to gravity,g = 9.8 m/s²

a) magnitude of mechanical energy lost

  Mechanical energy lost is equal top gain in kinetic energy

   ME_{Lost}=\dfrac{1}{2}mv^2

   ME_{Lost}=\dfrac{1}{2}\times 70\times 4.35^2

   ME_{Lost}=662.29\ J

b) Work done = Force x displacement

    W = F. x

     F = μ mg

    W = μ mg . x

Work done is equal to 662.29 J

  x=\dfrac{W}{\mu m g}

using the coefficient of the friction,μ = 0.7

  x=\dfrac{662.29}{ 0.7\times 70\times 9.8}

     x = 1.38 m

Hence, the runner will slide to 1.38 m.

3 0
3 years ago
a0.155 kg arrow is shot from ground level, upward at 31.4 m/s. what is its kinetic energy (ke) when it is 30.0 m above the groun
swat32

The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J

Assuming air friction is negligible,

a = - 9.8 m / s²

u = 31.4 m / s

s = 30 m

v² = u² + 2 a s

v² = 31.4² + ( 2 * - 9.8 * 30 )

v² = 985.96 - 588

v² = 397.96 m / s

KE = 1 / 2 m v²

KE = 1 / 2 * 0.155 * 397.96

KE = 0.0775 * 397.96

KE = 30.85 J

Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J

To know more about kinetic energy

brainly.com/question/24360064

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1 year ago
In a department store an airstream from a hose connected to the exhaust of a vacuum cleaner blows upward at an angle and support
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The air blows under the ball to support it.
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3 years ago
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