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3 years ago

A force of 40N is applied to a 28 g mass, what is the acceleration? (round to the hundredths place)

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

Answer:

1428.6m/s²

Explanation:

Given parameters:

Force applied on the body = 40N

Mass of the body = 28g

1000g = 1kg

28g will therefore be 0.028kg

Unknown:

Acceleration = ?

Solution:

To solve this problem, we use the expression derived from Newton's second law of motion.

Force = mass x acceleration

Insert the parameters and solve;

40 = 0.028 x acceleration

Acceleration = $\frac{40}{0.028}$ = 1428.6m/s²

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A ball is rolling horizontally at 3.00 meters per second as it leaves the edge of a tabletop 0.750 meter above the floor. The ba

NikAS [45]

Answer: 9.81m/s^2

Explanation: since it’s free fall and friction is neglected, the magnitude of the ball’s acceleration will be equal to the general acceleration due to gravity which is 9.81m/s^2

8 0

3 years ago

Read 2 more answers

Consider two point charges located on the x axis: one charge, q1= -11.0 nC, is located at x1= -1.675 m; the second charge, q2= 3

Mekhanik [1.2K]

Answer:

Please refer to the figure.

q1 is a negative charge, and q2 and q3 are positive charges. So, the force exerted by q1 on q3 is attractive, and the force exerted by q2 on q3 is repulsive, which means F13 is directed towards left, and F23 is also directed towards left.

$F_{13} = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_3}{r_1^2} = \frac{1}{4\pi \epsilon_0}\frac{11\times 10^{-9}\times 47.5\times10^{-9}}{(-1.675 - (-1.18))^2} = 1.92\times 10^{-5}N$

$F_{23} = \frac{1}{4\pi\epsilon_0}\frac{q_2q_3}{r_2^2} = \frac{1}{4\pi\epsilon_0}\frac{31\times 10^{-9} \times 47.5\times 10^{-9}}{(-1.18 - 0)^2} = 9.5\times 10^{-6}$

The net force on q3 is the sum of these two forces:

$F_{net} = F_{13} + F_{23} = -1.92\times 10^{-5} + (-19.5\times10^{-6}) = 2.8\times 10^{-5} N$

Since both forces are directed towards left, their sign should be negative.

5 0

3 years ago

What concept or principle best explains why

andreev551 [17]

Answer:

d. Newton's first law

I hope this helps you

5 0

3 years ago

Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)

stepladder [879]

Answer:

$m(t)=20e^{-0.5t}$

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope $(t_{1/2})$ = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

$m(t)=m_0e^{-kt}$

Where,

$m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year$

So, the equation is: $m(t)=20e^{-kt}$

At half-life, the mass is reduced to half of the initial value.

So, at $t=t_{1/2},m(t)=\frac{m_0}{2}$ . Plug in these values and solve for 'k'. This gives,

$\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5$

Hence, the equation for the mass remaining is given as:

$m(t)=20e^{-0.5t}$

8 0

3 years ago

An electric car has a battery that can hold 16 kwh of energy. if the battery is designed to operate at 340 v, how many coulombs

Zanzabum

charge needed= 169412 C

Explanation:

energy stored in the battery= 16 KWh= 16 (1000 W)(3600 s)= 5.76 x 10⁷ J

Energy stored is given by U= q V

q= charge

V= voltage= 340 V

so 5.76 x 10⁷ = q (340)

q= 169412 C

4 0

4 years ago