All categories

3 years ago

You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the

Physics

1 answer:

Sphinxa [80]3 years ago

4 0

Answer:

Option (c) : 20°C

Explanation:

$t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2}$

T(final) = 500* 10 + 100*70/600 = 20°C

You might be interested in

A proton is going 2,000 m/s into a magnetic field of 300 T. How much force does it feel. The charge of a proton = 1.602 x 10^-19

NemiM [27]

This depends on the direction of the velocity vector to the magnetic field vector. The force is F=q(VxB) ("x" is the cross product.) The max force is when V and B are perpendicular. Then F=qVB = (1.602e-19)(2000)(300) = 9.612e-14 N

7 0

3 years ago

the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0

3 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Pavlova-9 [17]

The answer is number two, number four, and number one

3 0

3 years ago

Read 2 more answers

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 49.

UkoKoshka [18]

Answer:

1.1397 Nm

Explanation:

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.

If the muscle generates a force

$F = 49.5 N$ and $r = 2.65 cm$ , then the torque is equal to $rF$

we see that r = 2.65 cm = 0.0265 m

therefore

torque = 0.0265 x 49.5

= 1.1397 Nm

4 0

3 years ago