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2 years ago

How long would a 2950 N force need to act to cause a 6000 kg delivery truck to accelerate from 4.0m/s to 29.0m/s?

Physics

1 answer:

AysviL [449]2 years ago

8 0

Answer:

Time, t = 50.8 seconds.

Explanation:

Given the following data;

Mass = 6000 kg

Force = 2950 N

Initial velocity = 4 m/s

Final velocity = 29 m/s

To find time;

First of all, we would determine the acceleration of the truck using the following formula;

Force = mass * acceleration

2950 = 6000 * acceleration

Acceleration = 2950/6000

Acceleration = 0.492 m/s²

Next, we find time using the first equation of motion;

$V = U + at$

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Making time, t the subject of formula, we have;

$t = \frac{V - U}{a}$

Substituting into the equation, we have;

$t = \frac{29 - 4}{0.492}$

$t = \frac{25}{0.492}$

Time, t = 50.8 seconds.

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What are the components of friction?

Tresset [83]

<h2>Answer:</h2>

Friction:

When an object slips on a surface, an opposing force acts between the tangent planes which acts in the opposite direction of motion. This opposing force is called Friction. Or in other words, Friction is the opposing force that opposes the motion between two surfaces.

The main component of friction are:

Normal Reaction (R):

Suppose a block is placed on a table in the above picture, which is in resting state, then two forces are acting on it at that time.

The first is due to its weight mg which is working from its center of gravity towards the vertical bottom.

The second one is superimposed vertically upwards by the table on the block, called the reaction force (P). This force passes through the center of gravity of the block.

Due to P = mg, the box is in equilibrium position on the table.

Coefficient of friction ( μ ):

The ratio of the force of friction and the reaction force is called the coefficient of friction.

Coefficient of friction, µ = force of friction / reaction force

μ = F / R

The coefficient of friction is volume less and dimensionless.

Its value is between 0 to 1.

Advantage and disadvantage from friction force:

The advantage of the force of friction is that due to friction, we can walk on the earth without slipping.
Brakes in all vehicles are due to the force of friction.
We can write on the board only because of the force of friction.
The disadvantage of this force is that due to friction, some parts of energy are lost in the machines and there is wear and tear on the machines.

How to reduce friction:

Using lubricants (oil or grease) in machines.
Friction can be reduced by using ball bearings etc.
Using a soap solution and powder.

4 0

3 years ago

Fill in the blanks for the following:

storchak [24]

Answer:

a. 4.21 moles

b. 478.6 m/s

c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank

Explanation:

Volume of container = 100.0 L

Temperature = 293 K

pressure = 1 atm = 1.01325 bar

number of moles n = ?

using the gas equation PV = nRT

n = PV/RT

R = 0.08206 L-atm- $mol^{-1}$ $K^{-1}$

Therefore,

n = (1.01325 x 100)/(0.08206 x 293)

n = 101.325/24.04 = 4.21 moles

The equation for root mean square velocity is

Vrms = $\sqrt{\frac{3RT}{M} }$

R = 8.314 J/mol-K

where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol

Vrms = $\sqrt{\frac{3*8.314*293}{0.0319} }$ = 478.6 m/s

For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship

$\frac{Voxy}{Vnit}$ = $\sqrt{\frac{Mnit}{Moxy} }$

where

Voxy = root mean square velocity of oxygen = 478.6 m/s

Vnit = root mean square velocity of nitrogen = ?

Moxy = Molar mass of oxygen = 31.9 g/mol

Mnit = Molar mass of nitrogen = 14.00 g/mol

$\frac{478.6}{Vnit}$ = $\sqrt{\frac{14.0}{31.9} }$

$\frac{478.6}{Vnit}$ = 0.66

Vnit = 0.66 x 478.6 = 315.876 m/s

the root mean square velocity of the oxygen gas is

478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank

6 0

2 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in

cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

$T=2\pi w$

But we know as well that w is given by,

$w=\sqrt{\frac{k}{m}}$

Replacing,

$w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}$

So we have that

$T=0.827s$

7 0

2 years ago

How does the amount of friction affect the sum of forces

iren [92.7K]

Answer:

<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>

Explanation:

<h3>mark as brainliast</h3>

6 0

2 years ago

How long would a 2950 N force need to act to cause a 6000 kg delivery truck to accelerate from 4.0m/s to 29.0m/s?​

How long would a 2950 N force need to act to cause a 6000 kg delivery truck to accelerate from 4.0m/s to 29.0m/s?