How to solve 56 equals 1/2 velocity squared

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago

What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg

scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

$F_c = F_g$

$\frac{mv^2}{r} = \frac{GmM}{r^2}$

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

$v^2 = \frac{GM}{r}$

At the same time we know that the period is equivalent in terms of the linear velocity to,

$T = \frac{2\pi}{\frac{v}{r}}$

$v = \frac{2\pi r}{T}$

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

$v^2 = \frac{GM}{r}$

$( \frac{2\pi r}{T})^2 = \frac{GM}{r}$

$T = \sqrt{\frac{4\pi^2 r^3}{GM}}$

Replacing we have,

$T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}$

$T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})$

$T= 1.74hour$

Therefore the correct answer is C.

7 0

3 years ago

A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio

Maurinko [17]

The answer is D hoped this helped

4 0

3 years ago

A marshmallow is dropped from a 5.71 meter high pedestrian bridge, and 0.921 seconds later, it lands right on the head of an uns

Natalka [10]

let the height of the person with marshmallow on her head be "h"

consider the motion of the marshmallow after it is dropped from bridge.

Y₀ = initial position of the marshmallow above the ground = 5.71 m

Y = final position of marshmallow on head of person = h

v₀ = initial velocity of the marshmallow = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

t = time of travel for marshmallow = 0.921 sec

Using the kinematics equation

Y = Y₀ + v₀ t + (0.5) a t²

inserting the values

h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²

h = 5.71 - 4.16

h = 1.55 m

5 0

3 years ago

How do organisms interact with each other?

satela [25.4K]

Answer:

They interact in a lot of ways, prey, predator, etc.

6 0

3 years ago