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3 years ago

How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?

Physics

2 answers:

Feliz [49]3 years ago

7 0

Answer:

2 Newtons

Explanation:

kozerog [31]3 years ago

5 0

F = ma, where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.

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Two steamrollers begin 115 mm apart and head toward each other, each at a constant speed of 1.10 m/sm/s . At the same instant, a

NeX [460]

Answer:

The distance of fly travel is 115.06 m.

Explanation:

Given that,

Distance = 115 mm

Speed = 1.10 m/s

Speed of fly = 2.20 m/s

We need to calculate the relative speed

Using formula of relative speed

$v=v_{1}+v_{2}$

Put the value into the formula

$v=1.10+1.10$

$v=2.20\ m/s$

We need to calculate the time for the two steamrollers to meet each other

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{115}{2.20}$

$t=52.3\ sec$

We need to calculate the distance of fly travel

Using formula of distance

$d=vt$

Put the value into the formula

$d=2.20\times52.3$

$d=115.06\ m$

Hence, The distance of fly travel is 115.06 m.

3 0

3 years ago

H. The length of the shadow is different in evening and in the day. Justify

Vinvika [58]

the shadows are exactly the same length in the morning as they are in the evening.

is so obvious it’s that when the sun is low you get long shadows and when the sun is up in the sky like in the noon the shadow is shorter.

3 0

3 years ago

Read 2 more answers

Newton concluded that some force had to act on the moon because

Dimas [21]

Think in gravity, the moon is orbiting the Earth because of gravitational pull/force. Without this gravitational pull, the moon will just move in a straight line (Newton's first law).

6 0

3 years ago

Which of the following correctly describes magnetic poles.

Mademuasel [1]

Answer:

The magnetic poles are the places on the earth's surface where the magnetic field force is acting perpendicularly to the surface. The North magnetic pole is where the field is acting straight down; at the South magnetic pole, it is acting straight up.

Explanation:

3 0

3 years ago

Two identical conducting spheres, having charges of opposite sign, attract each otherwith a force of 0.108 n when separated by 5

lisov135 [29]

Let's start from the final situation. After the two spheres are connected with the conducting wire, the total charge distributes equally between the two spheres (because they are identical). We can call the charge on each sphere $Q/2$ , with Q being the total charge.
The electrostatic force in this situation is 0.0360 N, so we can write
$F=k \frac{( \frac{Q}{2} )^2}{r^2}$
where k is the Coulomb's constant and $r=50.0 cm=0.50 m$ is the separation between the two spheres. Using F=0.0360 N, we can find the value of Q, the total charge shared between the two spheres:
$Q= \sqrt{ \frac{4Fr^2}{k} } = \sqrt{ \frac{4(0.0360 N)(0.50 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=2.0 \cdot 10^{-6}C$

Now let's go back to the initial situation, before the conducting wire was attached; in this situation, the two spheres have a charge of $q_1$ and $q_2$ , whose sum is Q:
$Q=q_1 + q_2$
The electrostatic force between the two spheres in the initial situation is:
$F=k \frac{q_1 q_2}{r^2}$
And since we know F=0.108 N, we find
$q_1 q_2 = \frac{Fr^2}{k} = \frac{(0.108 N)(0.50 m)^2}{8.99 \cdot 10^9 N m^2 C^{-2}}=3.0 \cdot 10^{-12} C$
But the problem tells us that the two spheres have charges of opposite sign, so we must put a negative sign:
$-3.0 \cdot 10^{-12} C = q_1 q_2$

So now we have basically a system of 2 equations:
$2.0 \cdot 10^{-6} C = Q = q_1 + q_2$
$-3.0 \cdot 10^{-12} C = q_1 q_2$
If we solve it, we find the initial charge on the two spheres:
$q_1 = -1 \cdot 10^{-6}C$
$q_2 = +3 \cdot 10^{-6 } C$

6 0

3 years ago