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3 years ago

When is an object moving in uniform circular motion?

Physics

1 answer:

Tems11 [23]3 years ago

4 0

Answer: An object undergoing uniform circular motion is moving

Explanation:

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How much heat will be needed to warm 187 grams of water from 10 0C to 90 0C?

kvv77 [185]

<h3>Hello there!</h3>

Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.

The equation commonly accepted to find the answer to questions like these is the specific heat formula.

The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.

The information given:

m = 187 grams

c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)

ΔT = 80 degrees

Now just plug everything in to solve.

Q = 187 * 4.184 * 80

Q = 62592.64

So you have your answer: 62592.64 joules.

Hope this helped!

5 0

3 years ago

How many significant figures does the number 91010.0 have?

kirza4 [7]

Answer:

6 (six)

Explanation:

5 0

3 years ago

Read 2 more answers

A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The

just olya [345]

Force is defined as the rate of change of momentum.
The initial amount of momentum is $mv$ because water stops when it hit the wall total change of momentum must be $\Delta p=mv$ .
Now let's calculate the force.
$F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v$
We need to find $\frac{dm}{dt}$ . This is the amount of water hiting the wall per second.
$\frac{dm}{dt}=\rho Av$
Our final formula would be:
$F=\rho Avv=\rho Av^2$
And now we can calculate the answer:
$F=1000\cdot5\cdot 10^{-4}\cdot(20)^2=200 N$

6 0

3 years ago

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se

Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3} \\\\$

The velocity of the particle is calculated as the change in the position of the particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$

4 0

2 years ago

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s

jok3333 [9.3K]

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.

6 0

3 years ago