The specific heat of aluminum is 900 J/(kg·°C). How much heat does it take to raise 1 kg of aluminum 4°C?

mestny [16]

Answer:

kftisgkstisirstizurzursrus

3 0

3 years ago

Elena pulls a sled with her younger brother and sister across the snow covered yard. Then she pulls the same sled across a smoot

mylen [45]

The answer is B.
Hope this helps!

5 0

3 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

3 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago

It's a snowy day and you're pulling a friend along a

NeX [460]

The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

<h3>How to find the coefficient of friction between the sled and the snow?</h3>

Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
To solve the problem, we have to draw the free body diagram of the given system.
We have given with the following values,

$a=0\\\alpha =35^0\\T=75N\\m=57kg$