1. Gwen exerts a 16 N horizontal force as she pulls a 32N sled across a cement

Gabriel is performing an experiment in which he is measuring the energy and work being done by a ball rolling down a hill. Which

melamori03 [73]

Answer:

They will both be recorded in Joules

Explanation:

- The energy of an object refers in general to the mechanical energy of the object, which is the sum of its kinetic energy (KE), which is the energy of motion, and of its potential energy (PE), which is the energy due to its position above the ground:

$E=PE+KE$

- The work done by an object is equal to the product between the force exerted by the object, F, and the displacement, d:

$W=Fd$ (1)

The SI units of the energy is the Joules, J. The basic equivalence with SI units can be seen from the definition of kinetic energy:

$KE=\frac{1}{2}mv^2$

where

m is the mass, measured in kilograms (kg)

v is the speed, measured in meters per second (m/s)

So we have, since the speed is squared

$1 J = 1 kg\cdot \frac{m^2}{s^2}$

Now we can check from eq.(1) that the work as the same units. In fact, the force is measured in Newtons (N), and we know that force is equal to mass times acceleration, so we have:

$1 N = 1 kg \cdot \frac{m}{s^2}$

Distance is measured in meters (m), so the work is measured in

$[W]=[kg \cdot \frac{m}{s^2}][m]=[kg\cdot \frac{m^2}{s^2}]$

Which corresponds to the definition of Joule. Therefore, both energy and work are measured in Joules.

6 0

4 years ago

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Part A : A cylindrical water tank 10cm high and 520cm in diameter is filled with solution with a density of 0.75 g/cm^3

lisov135 [29]

Answer: $0.1066\ psi, 15.611\ kN$

Explanation:

Given

the height of the tank is $h=10\ cm$

The diameter of the tank is $d=520\ cm$

Density of solution $\rho=0.75\ g/cm^3\ or\ 750\ kg/m^3$

$(a)$ Water pressure at the bottom of the tank is

$P=\rho gh\\P=750\times 9.8\times 0.1\\P=735\ Pa$

$1\ Pa=0.000145038\ psi$

$\Rightarrow P=735\ Pa\ or\ 0.1066\ psi$

$(b) \text{Average force on the bottom is the product of pressure and area of the base}$

$F_{avg}=735\times \pi \cdot (\frac{520}{200})^2\\\\F_{avg}=735\times 3.142\times 6.76\\F_{avg}=15,611.34\ N\ or\ 15.611\ kN$

6 0

3 years ago

A person on a diet loses 1.6 kg in a week. How many micrograms/second (µg/s) are lost?

Ivan

Based on the given values above, in order for us to get the answer, we need to convert the units first. So in 1 kilogram, there is 1,000,000 micrograms. In this case, 1.6 kilograms is 1,600,000 micrograms. For the week to seconds, 1 week is equivalent to 604,800 seconds. Therefore, 1,600,000 micrograms/604,800 seconds. So we are going to simplify this. So it would be 2.65<span>µg/s. Hope this answers your question.</span>

8 0

3 years ago

What acceleration will result when a 12 N net force applied to a 3 kg object?

geniusboy [140]

Answer:

4m/s

Explanation:

due to newtons second law of motion

the accelerations that result when a 12-N net force is applied to a 3-kg object. A 3-kg object experiences an acceleration of 4 m/s/s.

HOPE THIS HELPS PLEASE MARK AS BRAINLIEST:)

8 0

4 years ago

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An applied force of 20 N is used to accelerate an object to the right across a

vichka [17]

Answer:

$F_{norm} = 100 N$

$F_{net}=10 N$

$\mu = 0.10$

$m = 10 kg$

$a=1.0 m/s^2$

Explanation:

To determine the normal force, we just need to analyze the situation along the vertical direction.

The box along the vertical direction is in equilibrium, so the equation of the forces is

$F_{norm} - F_{grav} = 0$

which means that

$F_{norm} = F_{grav} = 100 N$

The net force can be determined by looking at the situation along the horizontal direction (since the net force in the vertical direction) is zero. Here we have:

- An applied force of 20 N forward, $F_{app} = 20 N$