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3 years ago

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In SI units, what is the magnitude the net force acting on a 1,152 kg car that accelerates uniformly along a straight line from

a speed of 3 m/s to a speed of 17 m/s in 5s?

1 answer:

geniusboy [140]3 years ago

7 0

Answer:

Fnet = 14515.2 Newton

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

$Fnet = Fapp + Fg$

Where;

Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.

Given the following data;

Mass = 1,152 kg

Initial velocity, u = 3m/s

Final velocity, v = 17m/s

Time, t = 5 seconds

To find the magnitude of the net force;

First of all, we would determine the acceleration of the car.

Acceleration = (v-u)/t

Acceleration = (17 - 3)/5

Acceleration = 14/5

Acceleration = 2.8m/s

To find the applied force;

Fapp = mass * acceleration

Fapp = 1,152 * 2.8

Fapp = 3225.6 N

Next, we would find the force exerted on the car due to gravity.

Fg = mass * acceleration due to gravity

We know that acceleration due to gravity is equal to 9.8m/s²

Fg = 1152 * 9.8

Fg = 11289.6N

Substituting the values into the equation, we have;

Fnet = 3225.6 + 11289.6

Fnet = 14515.2 Newton

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If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = $\frac{3}{1000}$ kg

Therefore F = 0.003×5 = 0.015 m/s2

3 0

2 years ago

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago

The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0

AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t represented by function :

$p(t)=t^{2} +2t-4$

Now,

When function p(t) = 11 , t will be

$p(t)=t^{2} +2t-4$

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0

t = 3 , -5

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = $\frac{position}{time}$

therefore to get the value of velocity from function p(t)

we have to differentiate the function with respect to time

$\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)$

v(t) = 2t + 2

where v(t) = velocity at that time

as t = 3 for p(t)=11

so ,

v(t) = 2t + 2

v(t) = 2*3 + 2

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know more about function here:

brainly.com/question/12431044

#SPJ4

4 0

2 years ago

A moving truck has more __ energy than a parked truck

Karo-lina-s [1.5K]

Kinetic energy than parked

7 0

3 years ago

Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci

bagirrra123 [75]

Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

$\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\$

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

$\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4} = 1.205-1.033\\\\\frac{1}{C_4} = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\$

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

8 0

3 years ago