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3 years ago

Does a stove have mechanical energy

1 answer:

6 0

-- Any object has gravitational potential energy relative to any place
lower than where the object is. The stove in the kitchen has potential
energy relative to the basement floor.

-- If an object is not moving, then it has no kinetic energy. The stove has
no kinetic energy unless you throw it or drop it out of a window.

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According to newton's second law of motion, how much force will be required to accelerate an object at the same rate if its mass

guajiro [1.7K]

For every action ther is an equal and opposite reaction.

equal in force, opposite in direction

5 0

3 years ago

Because so little can be found of the first rocks to form on

fgiga [73]

Answer:

B is the answer

Explanation:

0-/-<

7 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago

A ball is tossed into the air from height of 8 feet with an initial velocity of 8 feet per second. find the time r(in seconds) i

-Dominant- [34]

For this case we have the following equation:
-16t + 8t + 8 = 0
The first thing we must do is rewrite the equation correctly.
We have then:
-16t ^ 2 + 8t + 8 = 0
Solving the polynomial we have:
t1 = -1/2
t2 = 1
We discard the negative root because we want to find the time.
Answer:
it takes for the object to reach the ground about:
t = 1 second

3 0

3 years ago

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of

Simora [160]

Answer:

Approximately $1.5\; \rm kg$ . (Assuming that this table is level, and that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ .

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$ .

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$ .

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$ .

Pressure is the ratio between normal force and the area over which this force is applied.

$\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}$ .

Equivalently:

$(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})$ .

In this question, $\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}$ .

It was found that $(\text{contact Area}) = 0.075\; \rm m^{2}$ (assuming that the entire side of this book is in contact with the table.

Hence:

$\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}$ .

If that the table is level, this normal force would be equal to the weight of this book:

$\text{weight} = 15\; \rm N$ .

Assuming that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ . The mass of this book would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}$ .

4 0

2 years ago