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3 years ago

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A weight lifter is trying to lift a 1500-N weight but can apply a force of only 1200 N on the weight. One of his friends helps h

im lift it all the way. What force was applied to the weight by the weight lifter's friend?

1 answer:

aliya0001 [1]3 years ago

4 0

Answer:

300 N

Explanation:

We have three forces acting on the object:

- W = 1500 N, downward: the weight of the object

- F1 = 1200 N, upward: the force exerted by the first man

- F2, upward: the force exerted by the friend

In order to lift the object, the sum of the two forces F1 and F2 must be at least equal to the weight of the object W. Therefore we can write

$F_1 +F_2 = W$

And from this equation, we can find the magnitude of F2:

$F_2 = W-F_1 = 1500 N-1200 N=300 N$

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An object weighs 15N in air and 13N when completely

anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

<em>From Archimedes' principle,</em>

<em>Upthrust = lost in weight = weight of water displaced.</em>

<em>U = W₁ - W₂..................... Equation 1</em>

<em>Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.</em>

<em>Given: W₁ = 15 N, W₂ = 13 N</em>

<em>Substituting these values into equation 1</em>

<em>U = 15 - 13 </em>

<em>U = 2 N</em>

<em>But,</em>

<em>Weight of water displaced = mass of water displaced. × acceleration due to gravity.</em>

<em>Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2</em>

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

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3 years ago

Mickey walks Pluto 4 miles North to the dog park. Mickey finds beautiful

yanalaym [24]

Answer is 12 milesBecause yeah

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2 years ago

The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figu

abruzzese [7]

Answer:

$\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}$

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25$

$\sf \implies KE =3.4 \times 25$

$\sf \implies KE = 85 \: J$

8 0

3 years ago

Read 2 more answers

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago

PLS ANSWER FAST WILL GIVE BRAINLEST!!!

Kipish [7]

Answer:

2 Newtons

Explanation:

$F = ma$

Therefore, your mass would be 1kg and your acceleration would be 2m/s/s

Plug the numbers into the equation:

$(1kg)(2m/s/s)$

which will equal

$2 Newtons$

8 0

3 years ago