All categories

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Physics

1 answer:

Vlad [161]3 years ago

3 0

Answer: 44.57°C

Explanation:

The following can be deduced from the question:

Specific heat of water = 4.186 J/kg

From the question, we can infer that 625 × 4.186 joules of heat will be lost when there's a 1°C drop of water.

We then calculate the amount if degrees that it'll take to cool for 7.96 x 10⁴J. This will be:

= 7.96 × 10⁴ /(625 × 4.186)

= 79600/(625 x 4.186)

= 79600/2616.25

= 30.43°C

The final temperature will then be:

= 75.0°C - 30.43°C

= 44.57°C

You might be interested in

A train moves at constant velocity of 50km/h. how far will it move in 0.5h?

Kobotan [32]

The formula for getting the distance will be distance = speed x time
                                                                               D = S x T
   speed or velocity = 50km/h
   time = 0.5 h
the equation will be done directly because it's already in it's SI units
            distance = 50km/h x 0.5h
            hour cancels hour and the equation remains = 50km x 0.5
                               Ans = 25 km
the train will move 25 km far in 0.5h

7 0

3 years ago

What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs

Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration which is experienced by the bicycle given by

$a=v^{2}/R$

This acceleration is only due to static force which given as

$f=ma\\f=m(v^{2}/R )$

The maximum value of the static force is given as

$Fs_{max}=u_{s}F_{N}$

where

FN is normal force equal to mass*gravity

Therefore when the car is on the verge of sliding

$f=fs_{max}\\ m(v^{2}/R )=u_{s}mg$

Therefore the minimum radius should be found by the bicycle move without sliding

$v^{2}/R=u_{s}g\\ R=v^{2}/u_{s}g\\R=(8.2)^{2}/(0.435*9.8)\\R=15.773m$

8 0

4 years ago

(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT

Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 × $10^1^3$ J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>

D + T → He + n
The D-T fusion reaction results in a Helium (He) and neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59× $10^6$ ×1.6× $10^-^1^9$ ) ÷ ( 5.030×1.66× $10^-^2^7$ )

= 3.37× $10^1^4$ J/Kg

= 3.0× $10^9$ J/g

Rate of fuel burning in grams per hour = 3.0× $10^9$ × 3600

= 3.6×3.0× $10^1^2$

= 1.08 × $10^1^3$ J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

4 0

2 years ago

When mantle rocks near the radioactive core are heated, they become less dense than the cooler, upper mantle rocks. These warmer

avanturin [10]

C.convection cells is the right answer

4 0

3 years ago

Read 2 more answers

Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force

gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

$F=k\frac{q_1q_2}{r^2}$

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

$F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}$

$F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N$

Ft=0N

Hope this helps!!

5 0

3 years ago

Read 2 more answers