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3 years ago

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Whether or not the information on an online source is credible and factual, is up to the user to decide. Please select the best

answer from the choices provided T F

2 answers:

juin [17]3 years ago

6 0

Answer:

this statement is True

Explanation:

Alexxx [7]3 years ago

4 0

Answer:

t

Explanation:

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List possible limitations to scientific models

Andru [333]

(a). Models are not totally inclusive, that is they do not contain all the possible data about an object. (b). Approximations are normally used, since these are not exact, different results may be observed for real events and model events. (c). Loss of accuracy as a result of over simplification.

5 0

4 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

4 years ago

What is the force of gravity on a dog in a space suit that's running around on the moon? The dog's body has a mass of 20 kilogra

gulaghasi [49]

Answer:

32.4 N

Explanation:

The force of gravity acting on the dog is equal to its weight on the Moon, which is given by

$W=mg$

where:

m is the mass of the dog

g is the acceleration of gravity on the moon

In this problem, we have:

m = 20 kg

g = 1.62 m/s^2 (acceleration due to gravity on the moon)

Substituting numbers into the equation, we find

$W=(20 kg)(1.62 m/s^2)=32.4 N$

4 0

3 years ago

Read 2 more answers

The level of water in an olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2.00 meters deep) needs to be

Goshia [24]

<span>We need to start by finding the surface area of the pool.
50 meters multiplied by 25 meters gives us 1250 square meters.
1250 square meters multiplied by .065 (6.5 cm in meters) gives us a volume of 81.25 cubic meters of water that needs to be pumped out of the pool.

There are 1000 liters in a cubic meter so this is 81250 liters. Divide by 4.2 to find the number of seconds required to pump out this much water and we get 19345.2 seconds. This equals approximately 5.37 hours.</span>

4 0

3 years ago

An 1120 kg car traveling at 17 m/s is brought to a stop while skidding 40 m. Calculate the work done on the car by friction forc

sergij07 [2.7K]

Answer:

Approximately $(-1.6 \times 10^{5}\; \rm J)$ (assuming that the car was on level ground.)

Explanation:

When an object of mass $m$ is moving at a speed of $v$ , the kinetic energy of that object would be $(1/2) \, m \cdot v^{2}$ .

Initial kinetic energy of the car:

$\begin{aligned}&\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2}\times 1120\; \rm kg \times (17\; \rm m \cdot s^{-1})^{2} \\ &\approx 1.6\times 10^{5}\; \rm J \end{aligned}$ .

After the car comes to a stop, the kinetic energy of this car would be $0\; \rm J$ because the car would not be moving.

Change to the kinetic energy of the car: $\text{Final KE} - \text{Initial KE} = -1.6 \times 10^{5}\; \rm J$ .

If the car is traveling on level ground, friction would be the only force that contributed to this energy change. Hence:

$\text{Work of friction} = \text{Energy change} = -1.6\times 10^{5}\; \rm J$ .

The value of the work that friction did is negative. The reason is that at any instant before the car comes to a stop, friction would be exactly opposite to the direction of the movement of the car.

The work of a force on an object is the dot product of that force and the displacement of that object. The dot product of two vectors of opposite directions is negative. Hence, in this question, the work that friction did on the car would be negative because the friction vector would be opposite to the movement of the car.

6 0

3 years ago