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2 years ago

5

4 This giraffe walks at a speed of 2 meters per second Where will the giraffe be 5 seconds from now? (The trees are 5 meters apa

rt)
O A Photo A
OB. Photo B
OC Photo
OD Photo D

1 answer:

Lorico [155]2 years ago

6 0

Answer:

photo c

Explanation:

2 meters per sec times 5 seconds = 10 meters traveled. 10 meters is photo c

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What is the answer?

Andrews [41]

Answer:

6.26 m/s

Explanation:

Pretty slow.... the PE (Potential Energy) at 2m will be converted to KE (Kinetic Energy) at the bottom of the track (neglecting friction)

PE = KE

mgh = 1/2 mv^2 divide both sides of the equation by 'm'

gh = 1/2 v^2 multiply both sides by 2

2 gh = v^2 take sqrt of both sides

v = sqrt ( 2gh) = sqrt ( 2*9.81*2) = 6.26 m/s

3 0

2 years ago

A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

8 0

3 years ago

Read 2 more answers

You wish to produce an emf of 41.0 mV using an inductor whose inductance is 13.0 H. You start with a current of 1.50 mA through

Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

emf of the inductor, V = 41.0 mV

inductance of the inductor, L = 13 H

initial current in the inductor, I₀ = 1.5 mA

change in time, Δt = 2.6 s

The emf of the inductor is given by;

$V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 = \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA$

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

6 0

3 years ago

An inclinded plane makes work easier by increasing the distance and blank the force

iragen [17]

Hi! I believe your answer is decreasing. <u>An inclined plane makes work easier by decreasing the amount of effort force needed, but increases the distance</u>. I hope this helps you! Good luck and have a great day. ❤️✨

7 0

3 years ago

A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processe

SVETLANKA909090 [29]

Answer:

$T_s = 6.8 degree C$

Explanation:

As per thermal radiation we know that rate is heat radiation is given as

$\frac{dQ}{dt} = \sigma eA (T^4 - T_s^4)$

here we know that

T = 34 degree C = 307 K

$A = 1.38 m^2$

e = 0.557

$\sigma = 5.67 \times 10^{-8} W/m^2K^4$

$\frac{dQ}{dt} = 120 J/s$

now we have

$120 = (5.67 \times 10^{-8})(0.557)(1.38)(307^4 - T_s^4)$

$120 = (4.36 \times 10^{-8})(307^4 - T_s^4)$

$T_s = 279.8 K$

$T_s = 6.8 degree C$

5 0

3 years ago