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stich3 [128]
4 years ago
9

David drove the first 6 hours of his journey at 65 km/hour and the last 3 hours of his journey at 80 km/hour. how far is the who

le journey in Km?
Physics
1 answer:
Sidana [21]4 years ago
6 0
2434+4678-4567+56788
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A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e
Gala2k [10]

Explanation:

Given that,

Charge acting on the object, q=-4\ nC=-4\times 10^{-9}\ C

Force acting on the object, F=19\ nC=19\times 10^{-9}\ C (in downward direction)

(a) The electric force acting in the electric field is given by :

F=qE

E is the electric field

E=\dfrac{F}{q}

E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, q=1.6\times 10^{-19}\ C

The force acting on the proton is :

F=qE

F=1.6\times 10^{-19}\times 4.75

F=7.6\times 10^{-19}\ N

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0
3 years ago
For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
3 years ago
Read 2 more answers
Sug<br>. To repeat the experiment what is it?<br>​
statuscvo [17]

Answer:

....more foam

Explanation:

7 0
3 years ago
I need help on this question!!!
Irina-Kira [14]
2500
--------=3125
0.8

This can be can be viewed as compression due to the direction of the arrows.
7 0
3 years ago
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
alexgriva [62]

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

y \propto \frac{1}{x}

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3


5 0
3 years ago
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