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NikAS [45]
3 years ago
11

What is the acceleration rate of an object if the amount of force applied on the object is 75 Newtons and the object has a mass

of 15 kilogram?
Physics
1 answer:
Anestetic [448]3 years ago
6 0

Recall Newton's second law:

F = ma

F is the force exerted on the object.

m is the mass of the object.

a is the acceleration of the object.

Given values:

F = 75 N

m = 15 kg

Substitute the terms in the equation with the given values and solve for a:

75 = 15×a

<h3>a = m/s²</h3>
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The temperature of 2.0 g of helium is increased at constant volume by ΔT. What mass of oxygen can have its temperature increased
gavmur [86]

Answer:

m = 9.6 g

Explanation:

Thermal energy given to helium gas at constant volume is given as

Q = nC_v \Delta T

so here we have

C_v = \frac{3}{2}R

n = moles

n = \frac{2}{4} = 0.5

so we have

Q = \frac{3}{2}R(0.5)\Delta T

now we know that

for oxygen gas we have

C_v = \frac{5}{2}R

for same amount of heat we have

Q = nC_v \Delta T'

\frac{3}{2}R(0.5)\Delta T = \frac{m}{32} (\frac{5R}{2}) \Delta T

m = \frac{0.75 \times 32}{2.5}

m = 9.6 g

8 0
3 years ago
What force is required to accelerate a 1840 kg car from 4.77 m/s to 23.5 m/s,
neonofarm [45]
A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2

.:. The acceleration is 28.27 m/s^2
7 0
2 years ago
Tamiko it's creating a list of characteristics traditionally associated with males in American society. which current characteri
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3 0
3 years ago
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As temperature increases, what happens to the density of ocean water? A. changes unpredictably B. decreases C. does not change D
Lelu [443]
As the temperature of water increases, the density of water will decrease.
4 0
3 years ago
If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in
motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

F_{initial} = \frac{kq_1q_2}{r^2}

Here,

k = Coulomb's constant

q_{1,2} = Charge at each object

r = Distance between them

As the distance is doubled so,

F_{final} = \frac{kq_1q_2}{( 2r )^2}

F_{final} = \frac{ kq_1q_2}{ 4r^2}

F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}

F_{final} = \frac{1}{4} F_{initial}

\frac{F_{final}}{ F_{initial}} = \frac{1}{4}

Therefore the factor is 1/4

6 0
2 years ago
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