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3 years ago

Explain why the car reaches a top speed even though the thrust Force remains constant at 3500N

1 answer:

wel3 years ago

4 0

<span>Air resistance will eventually equal the thrust force, so there is no resultant force meaning the car will stay at the same speed and can no longer accelerate. </span>

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A football is thrown with an acceleration of 15 m/s^2 and a force of 13 N. What is its mass?

-Dominant- [34]

I believe it is b. Lmk if I’m wrong

7 0

3 years ago

Read 2 more answers

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri

Makovka662 [10]

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

$P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J$

The kinetic energy of the spring

$K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J$

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

$W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J$

The work done by friction is -0.4454 Joules

8 0

3 years ago

A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.<br><br>What is the friction force?

zmey [24]

Answer:

54.3N

Explanation:

The normal force is perpendicular to the slope, so:

Normal Force = cos(37.2)(9.8*65).......507.39N

F(friction)=mu*F(normal)

F(friction)=(0.107)(507.39)

F(friction)=54.3N

3 0

3 years ago

What is the

Len [333]

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

6 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

4 years ago