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shepuryov [24]
2 years ago
7

What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2

Physics
1 answer:
jasenka [17]2 years ago
4 0

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

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Marysya12 [62]

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Hope this helped : )

4 0
2 years ago
If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
Ann [662]
We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
8 0
3 years ago
What is the equivalent resistance for a series circuit with three resistors : 5.0 ohms, 2.0 ohms, and 12.0 ohms
vesna_86 [32]

Answer:19ohms

Explanation:

equivalent resistance=5+2+12

equivalent resistance=19ohms

8 0
2 years ago
A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of
Sholpan [36]
F=ma
where:
F - force
m - mass
a - acceleration 

We transform this formula to get a:
a= \frac{F}{m}
a=\frac{2}{10}\frac{N}{kg}=0.2\frac{m}{s^{2}}
4 0
3 years ago
Read 2 more answers
Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o
KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹  N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹  N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea =  +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹  N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb =  -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴  +  -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0
3 years ago
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