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velikii [3]
3 years ago
8

If the mass of an object is 5 kg, the gravitational field strength is 2 and the change in height is 5 m, what is the change in g

ravitational potential energy stored?
Physics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

50joules

Explanation:

See the explanation from the image I have provided above.

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A hockey puck slides across the ice with an initial velocity of 7.3 m/s .It has a decleration of 1.1 m/s and is traveling toward
sashaice [31]

Answer:

t = 6.63 s

Explanation:

Given that,

Initial velocity of the puck, u = 7.3 m/s

Deacceleration of the puck, a = -1.1 m/s²

Distance traveled, d = 5 m

We need to find the time the goalie have to stop the puck. Using equation of motion.

v = u +at

v = 0 (stops)

So,

u=-at\\\\t=\dfrac{u}{-a}\\\\t=\dfrac{7.3}{-(-1.1)}\\\\t=6.63\ s

So, the required time is 6.63 seconds.

6 0
3 years ago
If you lift a 16 kg child from the floor to her bed, which is 1 meter high, how much have you increased her potential energy.
Amanda [17]
GPE=mgh
GPE=16•9.8•1
6 0
3 years ago
One way to conserve energy is to replace incandescent light bulbs with compact fluorescent bulbs. The fluorescent bulb typically
NemiM [27]

Answer:

     cos to = $ 24

Explanation:

When replacing the bulb only 25% of the energy is used, therefore

          W = 0.25 100

          W = 25 W

Let's look for the energy in the life of the bulb

          E = 25 10⁻³ 12000

          E = 300 Kwh

now we can calculate the cost using a direct proportion rule.

          Cost = 0.08 300

          cos to = $ 24

4 0
3 years ago
Please help with Physics Circuits!
Zigmanuir [339]
1) Let's start by calculating the equivalent resistance of the three resistors in parallel, R_2, R_3, R_4:
\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}
From which we find
R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:
R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega
So, the correct answer is D) 


2) Let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega

And these are connected in series with a resistor of 10 \Omega, so the equivalent resistance of the circuit is
R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega

And by using Ohm's law we find the current in the circuit:
I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A
So, the correct answer is C).


3) Let' start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} =  \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}
From which we find
R_{23} = 2.5 \Omega
Then these are in series with all the other resistors, so the equivalent resistance of the circuit is
R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega

And by using Ohm's law we find the current flowing in the circuit:
I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:
V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V
So, the correct answer is D).


4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:
\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}
From which we find
R_{23} = 7.55 \Omega

Now all the resistors are in series, so the equivalent resistance of the circuit is:
R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega

The current in the circuit is given by Ohm's law
I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A

Now we can compare the voltage drops across the resistors. Resistor 1:
V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V
Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:
V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V
Resistor 4:
V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V

So, the greatest voltage drop is on resistor 1, so the correct answer is D).


5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is
P_3 = I_3^2 R_3 =  \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W
So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.
5 0
3 years ago
Read 2 more answers
A student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm. find the torque?
Anna71 [15]

Answer:

2N

Explanation:

Given that a student pulls down with a force of 40 N on a rope that winds around a pulley of radius 5 cm. find the torque?

Solution

Using the formula

Torque = Force × radius

I.e

Torque = F r

Where

F = 40N

r = 5 cm = 5/100 = 0.05m

Substitute all the parameters into the formula

Torque = 40 × 0.05

Torque = 2 N

Therefore, the torque on this scenario is 2 N

7 0
3 years ago
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