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3 years ago

An example of Newton's third law of motion is...

1 answer:

3 0

Newton’s third law states that for every action there is an equal and opposite reaction therefore, the third answer is right

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A glass of water sits on a scale. The scale reads 4.78 N. A steel bolt weighs 2.07 N. The density of steel is 7.86 g/cm3. We tie

miv72 [106K]

Answer:

The weight of water weight is 5.043 N.

Explanation:

Given that,

Force = 4.78 N

Weight of steel bolt = 2.07 N

Density of steel = 7.86 g/cm³

Suppose determine how much does the glass of water weight.

We need to calculate the volume of bolt

Using formula of density

$\rho_{b}=\dfrac{m}{V}$

$V=\dfrac{m}{\rho_{b}}$

Put the value into the formula

$V=\dfrac{2.07}{9.8\times7860}$

$V=2.6873\times10^{-5}\ m^3$

We need to calculate the force on bolt

Using buoyant force

$F=\rho gV$

Put the value into the formula

$F=1000\times9.8\times2.6873\times10^{-5}$

$F=0.263\ N$

We need to calculate the new weight of glass of water

Using formula for weight of glass of water

$W'_{w}=W_{w}+F_{b}$

Put the value into the formula

$W=4.78+0.263$

$W=5.043\ N$

Hence, The weight of water weight is 5.043 N.

4 0

3 years ago

How the volume of regular solids and of liquids is measured?

ahrayia [7]

Explanation:

the volume of solids is expressed in cubic measurememts of the length times the width, then times the weight.

5 0

3 years ago

Read 2 more answers

How can electric energy be converted to heat energy?

Goshia [24]

To do that, you must pass electric current through a substance
that electrons have to spend energy to pass through.
The substance will be one that gets warm and dissipates heat
when electric current flows through it.
We'll say that the substance has "resistance", which we can measure.
The amount of heat that appears when current flows through it
will be (current²)·(resistance).

A few examples of things used for that purpose:

-- resistors
-- burners on electric stoves
-- coils of resistor-wire in a toaster
-- aquarium heater
-- electric clothes iron
-- electric coffee pot
-- blow-dryer
-- electric hair-curling iron
-- skinny tungsten wire in a light-bulb .

5 0

3 years ago

Darw a velocity time graph for body moving with uniform acceleration

Troyanec [42]

Answer:

constant gradient rising

Explanation:

7 0

2 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago