A 6 kg table has a net force of 15 N acting on it. What is the magnitude of the acceleration of the table?

Renald completed only one trial of his experiment. what effect will this most likely have?

Tatiana [17]

Remember the rule of thumb that every person learns in Elementary Science? You must do multiple experiments in order to get to a conclusion. In order for a conclusion to be valid you must test the conclusion multiple times. You wouldn't want a doctor to just test an aspirin 1 time on 1 patient and say yes it works correct? No, you would want him/her to test on multiple patients in multiple settings and conditions so that when you take an aspirin you know that it will work for what you are taking it for. So ..... with all that being said.....Your answer is (A). The results are more likely to have errors.

6 0

3 years ago

A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly accord

pochemuha

Answer:

427.392 kJ

Explanation:

m = Mass of gas = 4.5 kg

Initial temperature = 200 K

Final temperature = 360 K

R = Mass specific gas constant = 296.8 J/kgK

$\gamma$ = Specific heat ratio = 1.5

Work done for a polytropic process is given by

$W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ$

The work input during the process is -427.392 kJ

3 0

3 years ago

What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21

GalinKa [24]

The correct answer to the question is: 1 N.

EXPLANATION:

As per the question, the spring constant or the force constant of the spring is given as k = 100 N/m.

The original length of the spring L = 20 cm.

The stretched length of the spring L'= 21 cm.

Hence, the change in length will be-

∆L = L' - L

= 21 cm - 20 cm

= 1 cm

= 0.01 m

We are asked to calculate the magnitude of force acting on the spring .

From Hooke's law, we know that the restoring force that acts on the spring is proportional to the distance .

Mathematically it can be written as -

F = - kx.

Here, k is the force constant.

x is the change in length due to compression or elongation.

The negative sign is due to the fact that it is opposite to the applied force.

Hence, the applied force on the spring is calculated as -

F = kx

= k × ∆L

= 100 N/ m × 0.01 m

= 1 N.

Hence, the force acting on the spring is 1 N.

8 0

4 years ago

A radio changes what energy to what energy

koban [17]

Electrical energy to sound energy

3 0

3 years ago

A cart loaded with bricks has a total mass of 20.4 kg and is pulled at constant speed by a rope. The rope is inclined at 26.1 ◦

Readme [11.4K]

Answer:

Normal force, N = 154.5 N

Explanation:

Given that,

Total mass of the cart, m = 20.4 kg

Angle of inclination, $\theta=26.1^{\circ}$

Distance moved, d = 20.1 m

The coefficient of kinetic friction between ground and cart is 0.6

The value of acceleration due to gravity, $g=9.8\ m/s^2$

Let N is the normal force exerted on the cart by the floor. It is given by :

$N=mg-T\ sin\theta$

$T\ cos\theta=\mu N$

$T\ cos\theta=\mu (mg-T\ sin\theta)$

$T\ cos(26.1)=0.6 (20.4\times 9.8-T\ sin(26.1))$

T = 103.23 N

So, the normal force is :

$N=20.4\times 9.8-103.23\ sin(26.1)$

N = 154.5 N

So, the normal force exerted on the cart by the floor is 154.5 N. Hence, this is the required solution.

4 0

3 years ago