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3 years ago

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Which of Newton's laws explains why your hands get red when you press them hard against a wall? A. Newton's law of gravity B. Ne

wton's first law of motion C. Newton's second law of motion D. Newton's third law of motion

2 answers:

Talja [164]3 years ago

7 0

The ball and the wall experience the same force.

irinina [24]3 years ago

4 0

Answer:

D

Explanation:

newtons 3rd law of motion

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What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

Pani-rosa [81]

charge must be equal to 5.74 ×10⁻⁵

In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

→ Fnet =0

→ mg = qE

substituting the values we get :

0.00345 × 9.81 = q × 590

→ q = 5.74 ×10⁻⁵

Hence the charge must be equal to 5.74 ×10⁻⁵.

Learn more about charges here:

brainly.com/question/26092261

# SPJ4

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1 year ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago

Consider eight,eight-cubic centimeter (8 cm3) sugar cubes stacked so that they form a single 2 x 2 x 2 cube. How does the surfac

In-s [12.5K]

To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.

Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.

To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.

To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.

Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.

Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.

The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192

We can further simplify this ratio:

96:192

48:96

24:48

12:24

6:12

3:6

1:2

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2 years ago

Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at

Karo-lina-s [1.5K]

Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

Speed will be = squreroot ( 2*g*L)

L is length of pendulum

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3 years ago

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A circle graph shows that one-fourth of the students in a particular school ride the bus every day. What is this equivalent meas

Aleonysh [2.5K]

Answer:

it's 1.57... radians and 90°

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2 years ago