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3 years ago

14

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200 WW electric im

mersion heater in 0.400 kgkg of water.

1 answer:

ivolga24 [154]3 years ago

7 0

Answer:

The heat is 115478.4 J.

Explanation:

Given that,

Mass of water = 0.400 kg

Power = 200 W

Suppose, we determine how much heat must be added to the water to raise its temperature from 20.0°C to 89.0°C?

We need to calculate the heat

Using formula of heat

$Q=mc\Delta T$

Where, m = mass of water

c = specific heat

Put the value into the formula

$Q=400\times4.184\times(89-20)$

$Q=115478.4\ J$

Hence, The heat is 115478.4 J.

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Two particles experience an attractive electric force of 5 N. If one of the charges triples and

Olin [163]

3.75 N

Explanation:

F = k(q)(q)/r^2 = 5 N

F' = k(q)(3q)/(2r)^2

= k×3(q)(q)/4r^2

= (3/4)[k(q)(q)/r^2]

= (3/4)F

= (3/4)(5 N)

= 3.75 N

8 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

zubka84 [21]

Answer:

$894.12\ \text{N}$

$284.013\ \text{J}$

Explanation:

U = Energy = 76 J

x = Displacement of spring = 0.17 m

k = Spring constant

Energy is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 76}{0.17^2}\\\Rightarrow k=5259.51\ \text{Nm}$

Force is given by

$F=kx\\\Rightarrow F=5259.51\times 0.17\\\Rightarrow F=894.12\ \text{N}$

The magnitude of force must you apply to hold the platform is $894.12\ \text{N}$ .

Now $x=0.17+0.2=0.37\ \text{m}$

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 5259.51\times 0.37^2\\\Rightarrow U=360.013\ \text{J}$

Additional energy

$360.013-76=284.013\ \text{J}$

8 0

2 years ago

Planet C has a tilt of zero degrees. What seasonal changes would be expected on this planet?

marshall27 [118]

The second statement is your answer:

No change in temperatures between seasons. Since tilt is the angle between the direction of positive pole and the normal to the orbital plane, seasons occur due to the tilt of axis. If there were to be no, or zero degree tilt, then there will be no season because of the lack in temperature changes. The amount of sunlight received remains same throughout the year which results in no temperature change

8 0

3 years ago

Read 2 more answers

( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration

Talja [164]

Answer:

$4.725 m/s^{2}$

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then $a=\frac {F}{m}$ where a is acceleration, F is force and m is mass

Also making mass the subject of the formula $m=\frac {F}{a}$

For $m1= \frac {F}{13.5}$ and $m2=\frac {F}{3.5}$ hence $F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}$

4 0

3 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(c) 6.38 × $10^{-26}$ N

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

3 years ago