A squirrel sitting on the ground starts to run with an acceleration of 5.1 m/s².
How fast is the squirrel going after 0.25 seconds?
O A. 20.4m/s
OB. 0.49 m/s
OC. 3.1 m/s
OD. 1.28 m/s

Answer:

B. Outside the nucleus.

Explanation:

Electrons orbit the nucleus of the atom.

Explanation:

it is almost zero .this is because the distance and the electrostatic force are inversely proportional

Typing in the commands and instructions and pressing enter will put a new command in the command prompt​.

This refers to a words or phrase which causes the computer to execute certain tasks or functions.

Typing the exact instruction and entering it will ensure that the chosen tasks are found in the command prompt.

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Answer:

D. s=3x+4y

Explanation:

The line is at the 3rd column in the 4th row.

The initial speed v,  the mass m should be given at the bottom, so that the string reaches a maximum angle is  2.16 m/s.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a small mass of m = 187 g hangs from a string of length l = 1.21 m. The string reaches a maximum angle of θ = 36.5.  The acceleration due to gravity, g = 9.8 m/s²

The vertical height of the point mass h = l- lcosθ

h = 1.21 - (1.21cos 36.5°)

h = 0.2373 m

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy  = Potential energy

1/2 mv²= mgh

Substituting the values, we have

1/2 x v² = 9.8 x  0.2373

v = sqrt [2 x9.8 x  0.2373]

v= 2.16 m/s

Thus, the initial speed v the mass m should be given at the bottom, so that the string reaches a maximum angle is 2.16 m/s.

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The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

55 = (50sin37)t + ¹/₂(9.8)t²

55 = 30.1t + 4.9t²

4.9t² + 30.1t - 55 = 0

Solve using formula method;

t = 1.47 s

X = vxt

where;

X = (50 x cos37) x 1.47

X = 58.7 m

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Answer:

1.1299 x 10^-8   second    

Explanation:

Period = 1 / f = 1 / (8.85 * 10^7) = 1.1299 x 01^-8  sec

The amount of time it will take to have 125 grams of erbium-165 left

in the sample is; C: 31.2 hours

The formula for amount of substance after decay is;

N(t) = N₀(¹/₂)^(t/t_¹/₂)

We are given;

Half life; t_¹/₂ = 10.4 hours

Initial amount; N₀ = 1000 g

Amount left; N(t) = 125 g

Thus;

125 = 1000 * (¹/₂)^(t/10.4)

125/1000 = (¹/₂)^(t/10.4)

In 0.125 = (t/10.4) In 0.5

-2.08 = -0.693(t/10.4)

t/10.4 = -2.08/-0.693

t = 10.4 * 3

t = 31.2 hours

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To deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

The power generated in a circuit is directly proportional to the current flowing in the circuit.

P = IV

where;

Voltage in the circuit also increases with increase in the current flowing in the circuit.

Thus, to deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.

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Answer:

F = G M m / R^2        gravitational force on satellite of mass m

Fma / Fmb = (ma / Ra^2) \ (mb / Rb^2) = (ma / mb) * (Rb / Ra)^2

Fma / Fmb = 1 * 4 = 4

The force on A is 4 times the force on B

Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

At pH = 2 the ratio of the concentration of the hydrogen to the hydroxide ion is 10¹⁰. The hydrogen ion concentration at pH, 2 is 10⁻² and the hydroxide ion is 10⁻¹².

The parameter of measuring the concentration of the hydroxide and the hydrogen ion in the solution in order to determine the basic and the acidic nature is known as pH.

The concentration of H⁺ ions is calculated as:

pH = -log [H⁺]

2 = -log [H⁺]

[H⁺] = 10⁻²

The concentration of OH⁻ ions is calculated as:

pH + pOH = 14

pOH = 14 - 2

pOH = 12

Solving further:

pOH = -log [OH⁻]

12 =-log [OH⁻]

[OH⁻] = 10⁻¹²

At a pH = 2, the ratio of hydrogen ions to hydroxide ions:

10⁻² ÷ 10⁻¹² = 10¹⁰

Therefore, at pH = 2 the ratio is 10¹⁰.

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The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

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[tex]\textsf {C. Force}[/tex]

[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]

By using an electric field, it is feasible to differentiate between these different forms of radiation.

A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.

The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.

Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.

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Answer:equal mass

Explanation:the mass of both is 5kg

Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec

b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The velocity in the x-direction;

[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]

Velocity in y-direction;

[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]

The resultant velocity is found as 1.12 × 10⁵ m/sec.

The time taken to reach the ground is found as;

[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]

Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.

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Answer:

a community event

Explanation:

The resultant of the two forces is about 170 N.

The term resultant force has to do with the single force that has the same magnitude and direction as two or more forces acting together.

In this case, the both forces are acting in the forward direction. This implies that the resultant force is 90 N  + 80 N = 170 N.

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The force, in newtons, exerted by each of the 10 braces is 2.135 x 10⁴ N.

The force is defined as the shear stress or pressure applied per unit area.

F = P/A

Given is a 17.0-m-high and 11.0-m-long wall and its bracing under construction are shown in the figure(attached). The force is exerted by each of the 10 braces, if a strong wind exerts a horizontal force of 655 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths.

Considering the pivot at the base of wall.

From the equilibrium of forces, we have

r₁ x Fwind = r₁ x Fbsinθ

Put the values, we get

Fb = Fwind /10sin35°.............(1)

The wind force is also given by

Fwind = Horizontal force or pressure x Area

Fwind = F/A x hw

           =655 x 17 x 11

F wind = 122,485 N

From equation (1), we have

Fb = Fwind /10sin35°

Fb =  122,485 /10sin35°

Fb = 21,354.6080 N

0r Fb = 2.135 x 10⁴ N

Thus, the force exerted on each of the 10 brace is 2.135 x 10⁴ N.

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Answer:

50J

Explanation:

Work done=Fd

=5×10

=50J

The type of power plant burns material to make electricity is Fossil fuel. The correct option is C.

Energy is the ability to do work.

There are different kind of energies like kinetic energy, potential energy, gravitational energy and  heat energy.

The energy obtained from burning different type of materials like Coal, gas and oil and produces heat.

Fossil fuel power plants burn coal or oil to create heat and generate steam to run turbines which generate electricity.

Thus, the correct option is C.

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Answer:

the answer is ambivert have a good fourth of July

Answer:

i believe the answer is ambivert

Answer:

Fritional force

Explanation:

Fritional force is a force that always opposes the motion of an object and is frequently ignored

C

Explanation:

Applied force, air resistance

Answer:

The rod is stretched because there is a larger gravitational force on the mass closest to mass M.

F = G M m / r^2 - G M m / (r + l)^2       the difference of the forces on m's

F = G M m [1 / r^2 - 1 / (r + l)^2]

[1 / r^2 - 1 / (r + l)^2] = (r^2 + 2 r l + l^2 - r^2 / [r^2 (r + l)^2]

= (2 r l + l^2) / [r^2 (r + l)^2]

if l is small compared to r = (2 r l ) / [r^2 (r + l)^2]

= (2 r l ) / [r^4 + 2 r^3 l ]

(a)The value of W for the given condition will be 237.3 N.

(b)The distance must the fulcrum be moved to restore balance will be 0.070 m.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

Weight is applied to the right end of the rod, W₁= 225 N

Rod's weight,W₂ = 255

W₃ is the Weight at "x" distance from the left end.

x1  is the location of W₁ with respect to the left end, where L = 1.80 m

m is the location of the rod's center of mass from the left end.

The position of W from the left end; x₃ = 50 cm = 0.50 m

Xcm is the position of the center of gravity of the rod at the fulcrum from the left end;

Given that the fulcrum is balanced in this situation, the rod's center of gravity will be there.

Xcm= L - 75 cm

Xcm=1.80 m - 0.75 m

Xcm= 1.05 m

The value of x₂

x₂= 1.80 m/2 = 0.90 m

Currently, the horizontal center of gravity is determined by

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]

Substitute the values we get;

[tex]\rm 1.05 = \fraca{ (225 \times 1.80 + 255\times 0.90 + W\times 0.50)}{(225 + 255 + W)}\\\\ W = \frac{ (225\times 1.80 + 255\times 0.90 - 480 \times 1.05)}{(1.05 - 0.50)}\\\\ W = 237.3 \ N[/tex]

After moving the fulcrum 20.0 cm to the right,

The location of the W from the left end;

x₃ = .50 cm + 20 cm

x₃  = 70 cm

x₃  = 0.70 m

W₁, W₂, W, x₁, and x₂ have fixed values. Therefore, based on these values, the new center of gravity will be:

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]

[tex]\rm X_{cm} = \frac{(225\times 1.80 + 255\times 0.90 + 237.3\times 0.70)}{(225 + 255 + 237.3)}\\\\ X_{cm} = 1.12 m[/tex]

Fulcrum must be shifted by;

Y = (1.12 - 1.05)

Y= 0.070 meters.

W has been moved in the appropriate direction, thus the fulcrum should likewise be moved in the appropriate direction.

Hence, the value of W for the given condition will be 237.3 N. and the distance must the fulcrum be moved to restore balance will be 0.070 m.

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The work done in pulling the sled is 445,930.2 Joules.

The acceleration in the x direction, assuming that friction is negligible, is  20.85 m/s².

Time taken to pull the sled 185 m is 4.21s.

Work done is equal to product of force applied and distance moved.

Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.

Work = Force x Distance x cos(angle)

W= 1225 x 385 x cos 19°

W = 445,930.2 Joules

Thus, the  work done in pulling the sled is 445,930.2 Joules

From the Newton's second law of motion, we have'

F = ma

acceleration, a = 1225cos19° / ( 545 /9.81)

a = 20.85 m/s²

Thus, the acceleration in the x direction is  20.85 m/s²

Using the second equation of motion, we get

s = ut+ 1/2 at²

Substitute the values, we have

185 m = 0 +1/2 x 20.85 x t²

t = 4.21 s

Thus, the time taken to pull the sled 185 is 4.21s.

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Answer:

Explanation:

There are different ways to investigate the factors that affect resistance. In this practical activity, it is important to:record the length of the wire accuratelymeasure and observe the potential difference and currentuse appropriate apparatus and methods to measure current and potential difference to work out the resistance

Answer:

(D)6 times

Explanation:

Hello !

Because when we see the equation of momentum P=mv, where p=momentum

m=mass of the object&

v=velocity of the object

And, p is directly propositional to both mass and velocity.

Thus, p=mv

Therefore, we can conclude by when the mass is tripled and the velocity is doubled the momentum is increased by 6 units.