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Sedbober [7]
3 years ago
14

Which statement accurately describes the charge of the nucleus of an atom?

Physics
2 answers:
Alenkasestr [34]3 years ago
5 0
<span>The answer is D.The charge of a nucleus can change from positive to negative. Hope this helps.</span>
fredd [130]3 years ago
4 0
The charge of a nucleus can change from positive to negative.
You might be interested in
An astronaut decides to perform an experiment to monitor how much weight he loses during his stay on the International Space Sta
kumpel [21]

Answer:

The first one is: His weight on the Earth before take-off and the weight after take-off back on Earth once he gets back should be recorded as his Independent variable and his dependent variable.

The second one is: If he gained the weight back that he had lost while on the trip then you should disregard them unless that was the weight he was when he weighed himself after he got back.

The Third one is: The mass of an object is the amount of matter it contains, regardless of its volume or any forces acting on it. … Gravity is a force that attracts objects toward the Earth. The weight of the object is defined as the force caused by gravity on a mass.

Explanation:

I took  the quiz earlier. Hope this Helps you.

4 0
3 years ago
The baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance, what is th
photoshop1234 [79]

Answer:

9.81 N

Explanation:

Given that a baseball weighs 1 kg. The pitcher throws the ball at a velocity of 10 m/s straight up. Neglecting air resistance. The maximum height reached will be calculated by using the formula

V^2 = U^2 - 2gH

U = 10 m/s

g = 9.8m/s^2

At maximum height, V = 0

Substitute u and g into the formula

0 = 10^2 - 2 × 9.8 × H

19.6H = 100

H = 100/19.6

H = 5.1 m

The Kinetic energy on the ball will be

K.E = 1/2mv^2

K.E = 1/2 × 1 × 10^2

K.E = 1/2 × 100

K.E = 50 J

But energy = work done

WD = Force × distance (height)

The force that acts on the baseball when it is HALF WAY to the top of the path will be

F × 5.1/2 = 50

F = 100/5.1

F = 19.61 N

The weight acting downward will be

W = mg

W = 1 × 9.8

W = 9.8 N

The net force acting on the ball will be

Net force = F - W

Net force = 19.61 - 9.8

Net force = 9.81 N

Therefore, the net force that acts on the baseball when it is HALF WAY to the top of the path is 9.81 N

4 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 64.7 N64.7 N , Ji
WARRIOR [948]

Answer:

(a) Magnitude of force is 262.51 N

(b) Angle with East direction is -14.75^{o}

Explanation:

Force by Jack in vector form

\overrightarrow F _1} = 64.7{\rm{ N}}\left( {\hat i} \right)  

Force by Jill in Vector form is given by

\begin{array}{c}\\{\overrightarrow F _2} = 86.5{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 86.5{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\widehat j} \right)\\\\ = 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

Force by Jane is

\begin{array}{c}\\{\overrightarrow F _3} = 181{\rm{ N }}\cos {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( {\hat i} \right) + 181{\rm{ N }}\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\left( { - \widehat j} \right)\\\\ = 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\end{array}

Net force is:

\overrightarrow F = {\overrightarrow F _1} + {\overrightarrow F _2} + {\overrightarrow F _3}

Hence

\begin{array}{c}\\\overrightarrow F = 64.7{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\hat i} \right) + 61.16{\rm{ N}}\left( {\widehat j} \right) + 128{\rm{ N}}\left( {\hat i} \right) + 128{\rm{ N}}\left( { - \widehat j} \right)\\\\ = 253.86{\rm{ N}}\left( {\hat i} \right) - 66.84{\rm{ N}}\left( {\widehat j} \right)\\\end{array}

The net force will be given by

F = \sqrt {{{\left( {{F_x}} \right)}^2} + {{\left( {{F_y}} \right)}^2}

Since F_{x}=253.86N and F_{y}=-66.84N

\begin{array}{c}\\F = \sqrt {{{\left( {253.86{\rm{ N}}} \right)}^2} + {{\left( { - 66.84{\rm{ N}}} \right)}^2}} \\\\ = {\bf{262.51 N}}\\\end{array}

The direction of net force is:

\theta = {\tan ^{ - 1}}\left {\frac{{{F_y}}}{{{F_x}}}}

Since F_{x}=253.86N and F_{y}=-66.84N  

\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{ - 66.84{\rm{ N}}}}{{253.86{\rm{ N}}}}} \right)\\\\ = {\tan ^{ - 1}}\left( { - 0.2633} \right)\\\\ = {\bf{ - 14}}{\bf{.7}}{{\bf{5}}^{\bf{o}}}\\\end{array}

The angle with East direction is -14.75^{o}

Net force exerted on the donkey is in the south-east direction. So, the angle of net force from the east direction is -14.75^{o} and it is 14.75^{o} from the south.

5 0
3 years ago
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if
irina [24]

Answer:

The balloon will collapse

Explanation:

When air is removed from the bell jar, the balloon will collapse if the internal pressure from the balloon does not balance the atmospheric pressure from the surroundings.

5 0
3 years ago
A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a
marshall27 [118]

Explanation:

The given data is as follows.

     mass (m) = 0.160 kg,            spring constant (k) = 8 n/m,

     Maximum speed (v_{m}) = 0.350 m/s

Formula for angular frequency is as follows.

          \omega = \sqrt{\frac{{k}{m}}

    \omega = \sqrt{\frac{{8}{0.160}}

    \omega = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

            \nu_{max} = A \omega

                  A = \frac{\nu}{\omega}

                      = \frac{0.35}{7.07}

                      = 0.05 m

Hence, value of amplitude is 0.05 m.

(b)   Displacement = 0.030 m

Formula for mechanical energy is as follows.

            M.E = \frac{1}{2}kA^{2}

Putting the values into the above formula as follows.

            M.E = \frac{1}{2}kA^{2}

                   = \frac{1}{2} \times 8 \times (0.05)^{2}

                   = 9.8 \times 10^{-3} Joule

For x = 0.03,

As,     P.E = \frac{1}{2} \times kx^{2}

                = \frac{1}{2} \times 8 \times (0.03)

                = 3.6 \times 10^{-3}

Hence, calculate the kinetic energy as follows.

            K.E = M.E - P.E

                  = (9.8 \times 10^{-3} - 3.6 \times 10^{-3}) J

                  = 6.2 \times 10^{-3} J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is 6.2 \times 10^{-3} J.

7 0
3 years ago
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