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Radda [10]
3 years ago
9

I though a=9.8 is in free fall?

Physics
2 answers:
Luda [366]3 years ago
8 0
A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). ... It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity.
Elina [12.6K]3 years ago
7 0
9.8m/s^2, i’m not sure what you are asking
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Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr
IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

P=\frac{V^2}{R}

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is P=5.0 W, and the resistance of the resistor is R=15 k\Omega = 15000 \Omega, then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V

(b) 1.6 W

In this case, we have:

R=9.0 k\Omega = 9000 \Omega is the resistance of the resistor

V=120 V is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

R_1 = 100 \Omega\\R_2 = 150 \Omega

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V

While the greatest potential difference allowed in the second resistor is

V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0
3 years ago
A recipe for a sweet tea calls for 3 cups of sugar when you place a sugar into the tea it doesn't dissolve. How could you use di
barxatty [35]

Answer: You could dissolve it by heating it back up, then just cooling it down again.

Hope that helps!

6 0
3 years ago
Read 2 more answers
Do rolling down a grassy hill has kinetic energy or potential energy
Ann [662]

If you're moving, then you have kinetic energy.

If you're not at the bottom yet, then you still have
some potential energy left.

4 0
4 years ago
Read 2 more answers
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
3 years ago
Describe each class of lever and explain to characteristics of each
Nataly [62]

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example:  a see-saw

-- Class II lever

The load is between the fulcrum and the effort.

The Mechanical Advantage is always greater than 1 .

Example:  a nut-cracker, a garlic press

-- Class III lever

The effort is between the fulcrum and the load.

The Mechanical Advantage is always less than 1 .

I can't think of an example right now.

8 0
3 years ago
Read 2 more answers
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