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3 years ago

15

A student studies how an objects mass and speed are related to its kinetic energy. The table shows the results for one part of t

he investigation.

1 answer:

Bumek [7]3 years ago

4 0

Answer:

I THINK

Explanation:

ITS

You might be interested in

What do you guys know about emotional intelligence? And how is it used in the real world?

tester [92]

Emotional Intelligence is our own abilities of being able to identify and regulate our own emotions. This is used by individuals in the real world in order to regulate crying in front of everyone for example. Calming ourselves down by breathing is another good example of regulating emotion which plays into emotional intelligence.

Hope this helps mate :D

7 0

2 years ago

3.5 g of a hydrocarbon fuel is burned in a vessel that contains 250. grams of water initially at 25.00 C. After the combustion,

kherson [118]

Answer:

3.) 463 J/g

Explanation:

Heat given by the fuel is used to increase the temperature of the water

so it is given as

$Q = ms\Delta T$

here we will have

m = 250 g

$\Delta T = 26.55 - 25$

now we also know that

$s = 4186$

so we have

$Q = (0.250)(4186)(26.55 - 25)$

$Q = 1622.075 J$

so 3.5 g fuel gives above energy

so per gram of fuel will have total energy given as

$Q = \frac{1622.075}{3.5}$

$Q = 463.45 J/g$

6 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

You use 42 J of energy to move a 6.0 N object in 2 seconds. How far did you move it?

bogdanovich [222]

22 that's how much I moved

7 0

3 years ago

An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the

garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = $\dfrac{150}{45 cos 50^0}$

t = 5.19 s

a) height of arrow

$s = u t +\dfrac{1}{2}gt^2$

$s = v sin \theta \times t+\dfrac{1}{2}gt^2$

$s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2$

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = $\dfrac{u}{g}=\dfrac{30.29}{9.81}$

= 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0

3 years ago