A cart maintains a constant velocity of 100 m/s for 10 seconds. during this interval its acceleration is

Which of these examples displays CONDUCTION based on the definition?

dangina [55]

A most probably because conduction transfer heat by movement of collisions of particles and movement of electrons within a body

3 0

3 years ago

An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be 80cm.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is 20 cm.

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

#teamtrees #PAW (Plant And Water)

I hope this helps!

8 0

3 years ago

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The moon Phobos orbits Mars

shepuryov [24]

$27.9816 \times 10^{3} s$ is the period of orbit.

Explanation: 

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

$\frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}$

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant $6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}$

Here, given data

$M=6.23 \times 10^{23} \mathrm{kg}$

$R=9.38 \times 10^{6} \mathrm{m}$

Substitute the given values, we get T as

$\frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}$

$T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}$

$T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}$

Taking square root, we get

$T=27.9816 \times 10^{3} s$

4 0

3 years ago

A 2.0-kilogram mass is located 3.0 meters above

leva [86]

The correct answer is: Option (3) 9.8 N/kg

Explanation:

According to Newton's Law of Gravitation:

$F_g = \frac{GmM}{R^2}$ --- (1)

Where G = Gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

m = Mass of the body = 2 kg

M = Mass of the Earth = 5.972 × 10²⁴ kg

R = Distance of the object from the center of the Earth = Radius of the Earth + Object's distance from the surface of the Earth = (6371 * 10³) + 3.0 = 6371003 m

Plug in the values in (1):

(1)=> $F_g = \frac{6.67408 * 10^{-11} * 2 * 5.972*10^{24}}{(6371003)^2} = 19.63$

Now that we have force strength at the location, we can use:

Force = mass * gravitational-field-strength

Plug in the values:

19.63 = 2.0 * gravitational-field-strength

gravitational-field-strength = 19.63/2 = 9.82 N/kg

Hence the correct answer is Option (3) 9.8 N/kg

4 0

4 years ago

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Prove that the period of a simple pendulum T=2×22/7×rootl/g

Rudik [331]

Answer:

Explanation:

Check the attachment.

By the definition of the simple harmonic motion,

The motion of a particle whose acceleration is

1. Always directed towards the center of the motion.

2. Proportional to the distance from center of the motion.

We get, a ∝ -x

where

a = acceleration

x = distance of the particle from center of the motion

- sign says acceleration is in opposite direction of motion

Check the other attachment for proof of a = -ω²x (or a = -ω²y)

x and y are both use to denote distance from center of the motion

Download pdf

pdf

4 0

3 years ago