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3 years ago

7

If two objects are close to each other and the mass of one object decreases, what happens to the force of gravity acting between

the objects?

2 answers:

Diano4ka-milaya [45]3 years ago

8 0

The force decreases possible falling I hope this helps (ps I answered before him)^

Yanka [14]3 years ago

4 0

The force decreases possible falling

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The moon Phobos orbits Mars

dlinn [17]

Answer:

The moon Phobos orbits Mars

(mass = 6.42 x 1023 kg) at a distance

of 9.38 x 106 m. What is its period of

orbit?

Explanation:

Answer: 27.9816 x 10^3 is the period of orbit

8 0

3 years ago

A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v

gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, $d_{\frac{1}{2}}$ .

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

$d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}$

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, $d_{\text{remain}}$ , in which the driver reduces the speed to 40km/hr is

$d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}$ .

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by $t_{\text{remian}}$ .

$t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}$ .

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

$\text{speed} \ = \ \displaystyle\frac{\Delta d}{\Delta t} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{(0.5 \ \text{hr} \ + \ 1.5 \ \text{hr})} \\ \\ \\ \text{speed} \ = \ \displaystyle\frac{100 \ \text{km}}{2 \ \text{hr}} \\ \\ \\ \text{speed} \ = \ 50 \ \text{km hr}^{-1}$

Therefore, the average speed of the car is 50 km/hr.

8 0

2 years ago

Which type of map would you most likely use to locate mineral deposits? A. topographic B. geologic C. satellite D. hazard

qwelly [4]

I think you would be using a topographic Map, So the answer should be A

3 0

3 years ago

A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The

Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

$\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2$

$Kx^2=mv^2$

$v=\sqrt{\dfrac{kx^2}{m}}$

By putting the values

$v=\sqrt{\dfrac{kx^2}{m}}$

$v=\sqrt{\dfrac{1300\times 0.264^2}{25}}$

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

$V_f^2=V_i^2-2gh$

By putting the values

$0^2=1.9^2-2\times 10\times h$

h=0.1805 m

h=18.05 cm

4 0

3 years ago

A student, standing on a scale in an elevator at rest, sees that his weight is 840 N. As the elevator rises, his weight increase

ololo11 [35]

As per FBD while its accelerating upwards

we can say that

$F_n - mg = ma$

here normal force is given as

$F_n = 1050 N$

$W = 840 N$

now mass is given as

$m(9.8) = 840$

$m = 85.7 kg$

now we will have

$1050 - 840 = 85.7 \times a$

$a = 2.45 m/s^2$

Now while accelerating downwards we can say by FBD

$mg - F_n = ma$

again plug in all values

$840 - 588 = 85.7 \times a$

$a = 2.94 m/s^2$

5 0

3 years ago