Ask question

Ask question

All categories

1 year ago

15

A 0. 25-kg toy car experiences a net force of 9. 0 n while slowing down to a complete stop. What is the magnitude of its acceler

ation?.

1 answer:

mezya [45]1 year ago

6 0

Answer:

a = 36 m/s

Explanation:

Given:

F = 9 N

m = 0.25 kg

a= ?

Use the equation:

F = ma (divide by m on both sides to get a alone)

a = F/m= 9/0.25 = 36 m/s

You might be interested in

Which of the following items best embodies the physical property of conductivity?

Sladkaya [172]

C is the correct answer

7 0

3 years ago

Read 2 more answers

Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res

damaskus [11]

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

3 0

3 years ago

Two forces, one four times as large as the other, pull in the same direction on a 10kg mass and impart to it an acceleration of

notsponge [240]

Answer:

The acceleration of the mass is 2 meters per square second.

Explanation:

By Newton's second law, we know that force ( $F$ ), measured in newtons, is the product of mass ( $m$ ), measured in kilograms, and net acceleration ( $a$ ), measured in meters per square second. That is:

$F = m\cdot a$ (1)

The initial force applied in the mass is:

$F = (10\,kg)\cdot \left(2.5\,\frac{m}{s^{2}} \right)$

$F = 25\,N$

In addition, we know that force is directly proportional to acceleration. If the smaller force is removed, then the initial force is reduced to $\frac{4}{5}$ of the initial force. The acceleration of the mass is:

$\frac{25\,N}{20\,N} = \frac{2.5\,\frac{m}{s^{2}} }{a}$

$a = 2\,\frac{m}{s^{2}}$

The acceleration of the mass is 2 meters per square second.

4 0

3 years ago

Electromagnetic waves all move at the same wave speed. which of these offer solutions to slowing them down select all that apply

Lelechka [254]

option c...have the waves move thru the vacuum or space

6 0

3 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago