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Basile [38]
3 years ago
5

What’s the answer to this question

Physics
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

space = 66.24 [m]

Explanation:

To solve this problem we must remember that the average speed is defined as the relationship between a space traveled over a certain time.

Av = \frac{space}{time}

where:

space [m]

Av = average velocity = 3.6 [m/s]

time = 18.4 [s]

space = 3.6*18.4\\space = 66.24 [m]

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Which of the following is an example of sustainable use of natural resources? Question 21 options: 1) clear-cutting a forest 2)
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4 is thebest anser as itleads with harwesting
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when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0
3 years ago
A man-made satellite orbits the earth in a circular orbit that has a radius of 32000 km. The mass of the Earth is 5.97e 24 kg. W
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Answer:

= 3521m/s

The tangential speed is approximately 3500 m/s.

Explanation:

F = m * v² ÷ r

Fg = (G * M * m) ÷ r²

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v = √( G * M ÷ r)

G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴

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G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶

v = √1.24 * 10⁷  

v = 3521.36m/s

The tangential speed is approximately 3500 m/s.

8 0
4 years ago
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Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c
PIT_PIT [208]
Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
E=k_e \frac{q}{r^2}
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C

While the electric field generated by the second charge is
E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C
4 0
3 years ago
Is gravitational pull strengthened by distance?
topjm [15]

I'm not sure but I think that if objects are closer (less distant), than the gravitational pull is stronger. I'm sorry if I couldn't really help I remember learning about this kind of.

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