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Trava [24]
3 years ago
8

To warm 2.0 l of tea (d = 1.01 g/ml; sp. heat = a cook places a 500 g block of stone at a temperature of 200f into the teapot. a

ssuming that the tea was initially at 72f, what is the final temperature of the tea in f?
a. 33
b. 91
c. 110
d. 140
e. 200
Physics
1 answer:
IRISSAK [1]3 years ago
4 0
Volume of tea V = 2.0L = 2000 mL density d = 1.01 g/ mL mass of tea m = V * d = 2000mL * 1.01g/mL = 2020 gWhen we assume that the tea was initially at 72, the final temperature of the tea in F is 91. 
The answer in this question is B. 91
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A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
jeka57 [31]

Answer:

<h3>The answer is 45 J</h3>

Explanation:

The work done by an object can be found by using the formula

<h3>workdone = force × distance</h3>

From the question

distance = 3 meters

force = 15 newtons

We have

workdone = 15 × 3

We have the final answer as

<h3>45 J</h3>

Hope this helps you

7 0
2 years ago
41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
A proton in a particle accelerator is traveling at a speed of 0.99c
Gwar [14]

A proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

<h3>What is speed of proton?</h3>

The speed of a proton is the rate at which a proton is moving through a given space.

The given speed of the proton is 0.99c

where;

  • c is speed of light

<h3>What is speed of light?</h3>

The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.

The value of speed of light in a vacuum is given as 3 x 10⁸ m/s.

The speed of the proton is calculated as follows;

v = 0.99 x 3 x 10⁸ m/s.

v = 2.97 x 10⁸ m/s.

Thus, a proton in a particle accelerator is traveling at a speed of 0.99c has a speed magnitude of 2.97 x 10⁸ m/s.

Learn more about speed of proton here: brainly.com/question/14663642

#SPJ1

8 0
1 year ago
What is the weight of a basketball with a mass of 0.5 kg
IgorLugansk [536]

W = 4.9N. The weight of a basketball with a mass of 0.5Kg is 4.9N.

The weight of an object is the force of gravity on the object and can be defined as the product of the mass by the acceleration of gravity, w = mg.

W = (0.5Kg)(9.8 m/s²) = 4.9N

6 0
3 years ago
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