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2 years ago

What force is needed to accelerate an object 5 m/s^2 if the object has a mass of 10kg?

1 answer:

4 0

It is important to pay attention to the fundamental units in the given information of physics problems; if you carefully arrange a system so the units cancel out appropriately, the answer is usually correct.

Force is mass x acceleration, and the units of force are (Kg/m·s²). In this system, a 10 Kg system is accelerated by 5 m/s². Simply multiply:
10 Kg x 5 m/s² = 50 Kg·m/s² = 50 Newtons

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You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of

Sunny_sXe [5.5K]

Answer:

$P = 1.090\,N$

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

$\Sigma F = P - W + F_{D}$

$\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0$

Now, the exerted force is:

$P = \rho \cdot V \cdot g - c\cdot v$

The volume of a sphere is:

$V = \frac{4\cdot \pi}{3}\cdot R^{3}$

$V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}$

$V = 1.149\times 10^{-5}\,m^{3}$

Lastly, the force is calculated:

$P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )$

$P = 1.090\,N$

5 0

2 years ago

A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The

Ber [7]

Answer:

1. $v=14.2259\ m.s^{-1}$

2. $F_T=25.8924\ N$

3. $\lambda=29.6373\ m$

Explanation:

Given:

mass of slinky, $m=0.87\ kg$

length of slinky, $L=6.8\ m$

amplitude of wave pulse, $A=0.23\ m$

time taken by the wave pulse to travel down the length, $t=0.478\ s$

frequency of wave pulse, $f=0.48\ Hz=0.48\ s^{-1}$

$\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel$

$v=\frac{6.8}{0.478}$

$v=14.2259\ m.s^{-1}$

Now, we find the linear mass density of the slinky.

$\mu=\frac{m}{L}$

$\mu=\frac{0.87}{6.8}\ kg.m^{-1}$

We have the relation involving the tension force as:

$v=\sqrt{\frac{F_T}{\mu} }$

$14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }$

$202.3774=F_T\times \frac{6.8}{0.87}$

$F_T=25.8924\ N$

We have the relation for wavelength as:

$\lambda=\frac{v}{f}$

$\lambda=\frac{14.2259}{0.48}$

$\lambda=29.6373\ m$

8 0

2 years ago

Which of the following would most likely cause a decrease in the quality supplied

soldier1979 [14.2K]

AWhich of the following would most likely cause a decrease in the quantity supplied? A decrease in price.

4 0

2 years ago

A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro

max2010maxim [7]

Answer:

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

$KE_{Total}=KE_{Translational}+KE_{Rotational}$

$KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2$

Here

$I=\frac{1}{2}m_{wheels}*r^2$ meaning the 4 wheels,

So replacing

$KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2$

So,

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}$

$\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}$

$\frac{KE_{Rotational}}{KE_{Total}} = 0.018$

3 0

3 years ago

Skid Marx set the world record for acceleration as he went from zero to 10 m/s in 3 seconds. How far did he travel in that time?

iren [92.7K]

Answer:

12000 miles per hour

Explanation:

If skid Marx went from zero to 10 miles in 3 seconds the his speed would have been 12000 miles per hour.

8 0

2 years ago