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Fudgin [204]
2 years ago
13

A student puts a piece of ice into a beaker of cold water. Which two statements are true. Please help

Physics
1 answer:
lesya692 [45]2 years ago
3 0

Answer:

A. Thermal energy will move from the water to the ice.

B. Thermal energy will move from air to the water.

C. Thermal energy will move from the ice to the air.

D. Thermal energy will move from the water to the air.

A and B are correct

Explanation:

Because thermal energy will move from the water to the ice

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Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Wri
Ede4ka [16]

Answer:

6.71 × 10^8 mi/hr

Explanation:

Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.

To write it in the metric system, it has to be converted into miles/hour.

We know that,

1 minute = 60 seconds

60 minutes = 1 hour

1 kilometer = 1000 meter

1 miles = 1.6 kilometer

Now,

= \frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}

= 1.08 × 10^12 m/ hr (meter/hour)

= \frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}

= 6.71 × 10^8 mi/hr (miles/hour)

Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.

5 0
3 years ago
Describe the parts of a lever. Include the following terms (fulcrum, resistance arm and effort arm).
yarga [219]

Answer:

hi here is your answer and this is a very important question.

Explanation:

A lever is a rigid bar with three parts: the fixed point around which the bar pivots is the fulcrum: the effort arm (in-lever arm) is the part of the lever to which force is applied; the resistance arm (out-lever arm) is the part that bears the load to be moved.

5 0
3 years ago
A 0.4-kg toy train car moving forward at 3 m/s collides with and sticks to a 0.8–kg toy car that is traveling in the opposite di
Gemiola [76]

Hey there!

Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.

0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s

–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s


So, the cars are traveling at -0.33 m/s in the direction of the second car.


Hope this helps


<em>Tobey</em>

4 0
2 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
2 years ago
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
antiseptic1488 [7]
Answer: -4000 kg • m/s

3 0
2 years ago
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