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ANEK [815]
3 years ago
8

What is the acceleration of an object with a mass of 15 kg and a coefficient of friction of 0.18

Physics
1 answer:
11111nata11111 [884]3 years ago
7 0

Answer:

a = 1.764m/s^2

Explanation:

By Newton's second law, the net force is F = ma.

The equation for friction is F(k) = F(n) * μ.

In this case, the normal force is simply F(n) = mg due to no other external forces being specified

F(n) = mg = 15kg * 9.8 m/s^2 =  147N.

F(k) = F(n) * μ = 147N * 0.18 = 26.46N.

Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.

Thus, F(net) = F(k) = ma

26.46N = 15kg * a

a = 1.764m/s^2

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Veronika [31]

Answer:

M g H = 1/2 M v^2       potential energy = kinetic energy

v^2 = 2 g H = 2 * 9.80 * 6 = 117.6 m/s^2

v = 10.8 m/s    

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6 0
1 year ago
Hi Guys.I was just wondering if given two specific heat capacities (In my case copper and water) do you add them both together o
timama [110]

Answer:

yes

Explanation:

using law of HC(heat capacity), which is

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  • energy H=MCQ

Where M is mass of substance,C is specific heat capacity, and Q is temperature change

In case of two substance

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8 0
2 years ago
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qaws [65]
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8 0
3 years ago
Calculate the electric field associated to an electric dipole for two charges separated 10-8 m with a dipole moment of 10-33 C m
Alex Ar [27]

Answer:

18 N/C

Explanation:

Given that:

Electric field constant, k = 9*10^9 N/c

Distance, r = 10^-8 m

Dipole moment, p = 10^-33

Using the relation for electric field due to dipole :

E = [2KP / r³]

E = (2 * (9*10^9) * 10^-33) ÷ (10^-8)^3

E = (18 * 10^9 * 10^-33) ÷ 10^-24

E = [18 * 10^(9-33)] ÷ 10^-24

E = (18 * 10^-24) / 10^-24

E = 18 * 10^-24+24

E = 18 * 10^0

E = 18 N/C

5 0
3 years ago
What is the speed of a commercial jet which travels form New York to Los Angeles (4800) in 6 hours
Colt1911 [192]

Answer:

Speed = 800km}/hr

Explanation:

Given

Distance = 4800km

Time = 6hr

Required

Determine the speed of the jet

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Speed = \frac{Distance}{Time}

Substitute 4800 km for Distance and 6hr for Time

Speed = \frac{4800km}{6hr}

Speed = 800km}/hr

<em>Hence, the speed of the commercial jet is 800km/hr</em>

8 0
2 years ago
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