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3 years ago

What net force is needed to increase the velocity of a 5 kg object by 3 m/s over the course of 2.5 seconds?

Physics

2 answers:

trasher [3.6K]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf 6 \ Newtons}}$

Explanation:

According to Newton's Second Law of Motion, force is the product of mass and acceleration. The mass of the object is 5 kilograms, but the acceleration is unknown.

We can find acceleration using the following formula:

$a= \frac{ \Delta v}{t}$

The velocity of the object increases by 3 meters per second. It accelerates in 2.5 seconds.

Δv= 3 m/s
t= 2.5 s

Substitute the values into the formula.

$a= \frac{3 \ m/s}{2.5 \ s}$

$a= 1.2 \ m/s/s = 1.2 \ m/s^2$

Now we know the mass and acceleration.

m= 5 kg
a= 1.2 m/s²

Substitute the values into the force formula.

$F=ma$

$F= (5 \ kg)(1.2 \ m/s^2)$

$F= 6 \ kg*m/s^2$

Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 6 kilogram meters per second squared is equal to 6 Newtons.

$F= 6 \ N$

mixer [17]3 years ago

6 0

Hi there!

Recall Newton's Second Law:

$\large\boxed{\Sigma F = ma}$

∑F = net force (N)

m = mass (kg)

a = acceleration (m/s²)

We must begin by solving for the acceleration using the following:

a = Δv/t

In this instance:

Δv = 3 m/s

t = 2.5 sec

a = 3/2.5 = 1.2 m/s²

Now, plug this value along with the mass into the equation for net force:

$\Sigma F = 5(1.2) = \boxed{6 N}}$

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

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3 years ago

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?

aniked [119]

An applied force is a force that is applied to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
Normal force is the component, perpendicular to the surface (surface being a plane) of contact.
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Solution A.

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We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.

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