A truck is traveling at 80 m/s and has 12,000 J of kinetic energy. What is the truck’s mass?

Is baking a cake conduction, convection, or radiation?

Stolb23 [73]

Answer:

baking- is a method of cooking food that uses prolonged dry heat, normally in an oven, but also in hot ashes, or on hot stones. The primary source of heat is radiation. Fans within the oven will increase cooking times via convection of the air.

5 0

2 years ago

Which statement best describes how work and power are different? a. To find work we need to know force and distance; to find pow

weqwewe [10]

A. To find work we need to know F and S; to find power we need to know F and V

6 0

2 years ago

If the average time it takes for the cart from point 1 to point 2 is 0.2 s, calculate the angle θ from the horizontal of the tra

inn [45]

Answer:

hehe

Explanation:

I dont know because I am a noob ant study

5 0

2 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.

Serga [27]

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

$A=50*20=1000[cm^{2} ]$

But we must convert from square centimeters to square meters.

$1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]$

And the pressure is:

$P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]$

6 0

2 years ago