How much force is required to accelerate a 2 kg mass at 3 m/s2

Saturn exerts a much greater force on the moon Pan than Pan exerts on saturn. true or false

shtirl [24]

Answer: False

Explanation:

because according to universal law of gravitation, every object attracts every other object in the universe.

according to the newton's third law, for every action there is equal and opposite reaction.

So, both the saturn and pan exerts same force on each other

But from the newton's second law we have, F = ma & a = F/m

so, from the above equation acceleration produced by pan is high compared to the acceleration produced by saturn.

Therefore, acceleration produced by saturn is lower than the acceleration produced by pan

but force exerted by saturn is equal to the force exerted by pan

So, the answer is false

Hope it helped u,

pls mark as the brainliest

^_^

5 0

3 years ago

A group of students is investigating whether the smoothness of surface affects the force of friction. Each student pushes a book

Pani-rosa [81]

To find the average (aka mean) of a group of numbers, all we need to do is add them up then divide that number by how many numbers we started with

23 + 21 + 24 + 22 = 90

90 ÷ 4 = 22.5

Therefore, the average distance the book traveled on ice is 22.5 cm

15 + 18 + 16 + 17 = 66

66 ÷ 4 = 16.5

Therefore, the average distance the book traveled on concrete is 16.5 cm

In conclusion, the average distance the book traveled on ice is greater than the average distance the book traveled on concrete

Hope this helps you

-AaronWiseIsBae

5 0

4 years ago

Read 2 more answers

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

4 years ago

A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

qaws [65]

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

4 0

3 years ago

0. The volume of a cube is 27cm3. Find the length of edge of cube.

Dennis_Churaev [7]

Answer:

27 cm3

Explanation:

7 0

3 years ago