If 25.0 g of water at 21c is mixed with 45.0 g of water at 75c, what is the final temperature of the mixture?

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of

Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0

3 years ago

The atomic number, or ________ number, is the described as the number of _________ in the nucleus of an chemical element. A) pro

Alenkinab [10]

i think it is a and c idk

4 0

3 years ago

Read 2 more answers

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0

3 years ago

The intensity of light from a central source varies inversely as the square of the distance. If you lived on a planet only half

sertanlavr [38]

Answer:

the intensity will be 4 times that of the earth.

Explanation:

let us assume the following:

intensity of light on earth =J

distance of earth from sun = d

intensity of light on other planet = K

distance of other planet from sun = $\frac{d}{2}$ (from the question, the planet is half as far from the sun as earth)

from the question the intensity is inversely proportional to the square of the distance, hence

intensity on earth : J = $\frac{1}{d^{2} }$

J $d^{2}$ = 1 ... equation 1

intensity on other planet : K = $\frac{1}{(\frac{d}{2}) ^{2} }$ (the planet is half as far from the sun as earth)

K $(\frac{d}{2}) ^{2}$ = 1 ....equation 2

equating both equation 1 and 2 we have

J $d^{2}$ = K $(\frac{d}{2}) ^{2}$

J $d^{2}$ = K $\frac{d^{2}}{4}$

J = $\frac{K}{4}$

K = 4J

intensity of light on other planet (K) = 4 times intensity of light on earth (J)

5 0

3 years ago

In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a

fgiga [73]

Answer:

$S = 20903.4$ ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium equation would be

$F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59$

Let the displacement be $S_0$

As per newton's third law of motion

$v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} = 117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} = 73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 = 2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4$

3 0

3 years ago