Suppose a fast-pitch softball player does a windmill pitch, moving her hand through a circular arc with her arm straight. She releases the ball at a speed of 21.3 m/s21.3 m/s (about 47.6 mph47.6 mph ). Just before the ball leaves her hand, the ball's radial acceleration is 871 m/s2871 m/s2 . What is the length of her arm from the pivot point at her shoulder?

Answer:

3.0 x 10 and the exponent on ten is 5.

Explanation:

You are moving the decimal point over to the left 5 times, making the exponent a positive 5 and then you would put a decimal after 3. Making it 3.0 x 10, and the exponent 5.

The speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.

The given parameters;

speed of the x-ray, v = 300,000 m/s

The standard form of the given number represents the scientific notation of the number.

In scientific notation the speed of the x-ray can be expressed as follows;

300,000 m/s = 3.0 x 10⁶ m/s

Thus, the speed of x-rays in scientific notation is 3.0 x 10⁶ m/s.

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Answer:

plz give the question... floor the proper answer

Explanation:

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thank

me and follow ma

Answer:

240 meters

Explanation:

He is walking 2 meters per second, and there are 60 seconds per minute. There are 120 seconds in two minutes. So, 120 x 2 = 240. That's your answer!

Answer:

Explanation:

If a positive charge enters a magnetic field at 90 degrees the charge is deflected in a circular path by a force that acts perpendicular to it in line with Flemings right-hand rule

to derive the radius of the path of the charge we apply

F= mv^2/r=Bev

where

m= mass of the electronic charge

e=charge

B=magnetic field

v=average speed

r=radius

rearranging we have

r=mv^2/Bev

r=mv/Be

Explanation:

(a) The car will leave tangentially to the road, so it will take path d.

(b) Sum of forces on the car in the vertical direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces on the car in the centripetal direction:

∑F = ma

Nμ = m v²/r

mgμ = m v²/r

gμ = v²/r

v = √(grμ)

The maximum velocity is independent of mass, so adding sandbags will not work.

Answer:

C

Explanation:

The 15N object has more mass than the 10N object.

A 15 N object has more mass than the 10 N object due to their weight differences.

Weight is defined as the force exerted by the earth on an object on the surface of earth. The direction of this force is directed towards the center of earth. Mathematically -

W = mg

Where -

m is the mass of body

g is the acceleration due to gravity

Weight can also be written as -

W = ρVg     [ρ is density and V is volume]

Given is a person who dropped two objects from the same height. One object weighs 15 N and the other weighs 10 N.

We can write -

Weight of Object 1 [W1] = 15 N

Weight of Object 2 [W2] = 10N

We know that -

W = m x g

We can write it as -

m = W/g

Therefore -

Mass of Object 1 [M1] = 15/9.8 = 1.53 Kg

Mass of Object 2 [M2] = 10/9.8 = 1.02 Kg

Clearly -

M1 > M2

So, the 15 N object has more mass than the 10 N object.

Therefore, a 15 N object has more mass than the 10 N object due to their weight differences.

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Answer:

The work to be done is 638,761.12 J

Explanation:

Given;

mass of the car, m = 1150 kg

velocity of the car, v = 120 km/h = 33.33 m/s

Apply work energy theorem;

W = K.E

where;

W is the work done

K.E is the kinetic energy

W = ¹/₂mv²

W =  ¹/₂ (1150)(33.33)²

W = 638,761.12 J

Therefore, the work to be done is 638,761.12 J

Given:-

To Find: Speed of it.

We know,

v = s/t

where,

Putting the values,

v = (20 m)/(4 s)

→ v = 5 m/s (Ans.)

The speed of an object will be "5 m/s".

The given values are:

Distance,

Time,

As we know, the formula:

→ [tex]Distance (d)= Speed (v)\times Time (t)[/tex]

or,

→ [tex]Speed (v) = \frac{Distance (d)}{Time (t)}[/tex]

By putting the values, we get

→                [tex]= \frac{20}{4}[/tex]

→                [tex]= 5 \ m/s[/tex]

Thus the above answer is correct.

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Answer:

mass of the block is 7.2Kg

Tension on string A = 72N

Tension on the string with an angle is =  108N

Explanation:

The Horizontal string will have the cos component of the force...

Hence, mg cos60 = 36

by keeping acceleration due to gravity = 10 m/s^2, we get,

m = 7.2Kg

The tension in String A = mg = 7.2*10 = 72N

since the whole string holding the block is pulled by another string with 36 N

the resultant force will be the tension of the string with an angle.

Hence, 72 +36 = 108 N

Please let me know whether I got this right or not...

Answer:

Open and globular

Explanation:

Answer:

Explanation:

Let the internal resistance be r .

Since in open circuit the volt is 1.55 V , this will be the source voltage .

Source voltage = 1.55

If external resistance be R .

1.55 / (R + r ) = .500

R + r = 3.1 ohm

So sum of internal resistance and external resistance will be 3.1 ohm.

Answer:

F = 1.6 N

Explanation:

Newton's Second Law of Motion states that whenever an unbalanced force is applied on a body. It produces an acceleration in the body, in its own direction. The magnitude of the force is given by the following formula:

F = ma

where,

F = Force Required to Move the Object Upward = ?

m = mass of object

a = upward acceleration = 2 m/s²

but, we know that weight of object is given as:

Weight = mg

8 N = m(9.8 m/s²)

m = 0.8 kg

using values in the equation:

F = (0.8 kg)(2 m/s²)

F = 1.6 N

Answer:

18*10^10 meters

Explanation:

V= d/t         10 mins = 600 seconds

3*10^8 = d/600s

(3*10^8)*(6*10^2) = d

d = 18*10^10 m

Answer:

16.7m/s

Explanation:

Using v= u+ at

V = final velocity. U = initial velocity

So since the two movements are the same they will return with the same speed they went up

Then final velocity v= 0

So

0:= u +-9.8 x 1.7

U= 16.7m/s

Answer:

is there an equasion it gives you?

Explanation:

need a little more info.

Answer:

The wavelength of the wave is 1 m

Explanation:

Given;

mass of the string, m = 20 g = 0.02 kg

length of the string, L = 3.2 m

tension on the string, T = 2.5 N

the frequency of the wave, f = 20 Hz

The velocity of the wave is given by;

[tex]v = \sqrt\frac{T}{\mu} {}[/tex]

where;

μ is mass per unit length = 0.02 kg / 3.2 m

μ = 6.25 x 10⁻³ kg/m

[tex]v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{2.5}{6.25*10^{-3}} } \\\\v = 20 \ m/s[/tex]

The wavelength of the wave is given by;

λ = v / f

λ = (20 m/s )/ (20 Hz)

λ = 1 m

Therefore, the wavelength of the wave is 1 m

Answer:

towards the center of the circle

(which appears to be the last option in your list of possible answers)

Explanation:

The centripetal force points always towards the center of the circle described by the object moving.

The direction of centripetal force acting on an object is always towards the center of the circle in which the object is moving.

The given problem is based on the concept and fundamentals of centripetal force. When an object is known to move around a circular path, then there is a force comes to play acting towards the centre of circular path, known as centripetal force.

Thus, we can conclude that the direction of centripetal force acting on an object is always toward the center of the circle in which the object is moving.

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Answer:

c

Explanation:

Answer:

the answer is c i just did it

Explanation:

Answer:

1. a = 9.8 m/s²

2. t = 4.28 s

3. Vf = 42 m/s

4. Vf = 28 m/s  

Explanation:

1.

Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.

a = 9.8 m/s²

2.

The time taken by the ball to reach the ground can be calculated by using second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height = 90 m

Vi = initial velocity = 0 m/s

t = time taken = ?

Therefore,

90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²

t = √(18.36 s²)

t = 4.28 s

3.

In order to find final velocity we use first equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(4.28 s)

Vf = 42 m/s

4.

when the ball is at height of 50 m, it means it has covered:

h = 90 m - 50 m = 40 m

we use third equation of motion at this point:

2gh = Vf² - Vi²

(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²

Vf = √(784 m²/s²)

Vf = 28 m/s

Answer:

[tex]V_1=2mL[/tex]

Explanation:

Hello,

In this case, considering this as a dilution problem, the first step here is to consider that the moles of tetrodotoxin remains the same and just the volumes and concentrations are modified from the initial stock (1) and the required dilution (2):

[tex]V_1M_1=V_2M_2[/tex]

Whereas V is referred to volume and M to molar concentration, in this case in mM. In such a way, solving the V1 as the volume of the 3-mM solution we obtain:

[tex]V_1=\frac{V_2M_2}{M_1} \\\\V_1=\frac{6mL*1mM}{3mM} \\\\V_1=2mL[/tex]

It means you need 2 mL of the 3-mM solution.

Best regards.

Answer: a. Stall AOA increases, CL max increases.

Explanation:

Deploying slats or slots on an airfoil affect CL max and Stall AOA by increasing Stall AOA and also increasing CL max.

It should be noted that the coefficient of lift CL would rise through the use of slots due to increase in boundary later energy. The slots also leads to delay of stall by through increase in AOA.

Answer:

Resultant = 13km

Direction = 67.38° East of North

Explanation:

Given the following :

5km North ; 12km East

Resultant Displacement (r) :

r² = 5² + 12²

r² = 25 + 144

r² = 169

r = √169

r = 13

Direction:

Tangent = opposite / Adjacent

Tanθ = opposite / Adjacent

Opposite = 12 ; adjacent = 5

Tanθ = (12/5)

Tanθ = 2.4

θ = tan^-1(2.4)

θ = 67.38° east of north

Momentum = (mass) x (speed)

We don't know the mass of the bowling ball, but we know how much it weighs.  So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass.  The question doesn't tell us what planet the bowling ball is on.  So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.

But I urgently need the points, so I'm going to go out on a limb here and make a big assumption:  I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration  of gravity is around  9.8 m/s²  everywhere on the planet's surface.  NOW I can go ahead and answer the question that I just invented.

weight = m  g

Mass = (weight) / (g)

Mass = (78.0 N) / (9.8 m/s²)

Mass = 7.96 kilograms

Momentum = (mass) x (speed)

Momentum = (7.96 kg) x (8.00 m/s)

Momentum = 63.7 kg-m/s

Explanation:

Given:

Δx = 50 m

v₀ = 0 m/s

a = 0.2 m/s²

Find: v

The problem does not involve time.  Therefore, the required equation is:

v² = v₀² + 2aΔx

Plug in values and solve:

v² = (0 m/s)² + 2 (0.2 m/s²) (50 m)

v = 4.47 m/s

The velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.

To find the velocity of the car when it has moved a distance of 50 meters, we can use the following equation of motion:

v² = u² + 2as

Where:

v = final velocity

u = initial velocity (which is 0 in this case as the car starts from rest)

a = acceleration

s = distance traveled

Plugging in the values into the equation:

v² = 0² + 2(0.2)(50)

v² = 0 + 20

v² = 20

Taking the square root of both sides:

v = √20

v = 4.47 m/s

Therefore, the velocity of the car when it has moved a distance of 50 meters is approximately 4.47 m/s.

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Answer:

A

Explanation:

Answer:

the full segment is:  30 units long

which coincides with answer C in your list

Explanation:

If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:

RM = MS (equality among the two parts of the divided segment)

replacing RM with "5 x" and MS with "x + 12", we get:

5 x = x + 12

solving for x:

5 x - x = 12

4 x = 12

then x = 3

With this info, we can calculate the length of each half of the segment and consequently its full length:

RM = 5 x = 5 (3) = 15

then the full segment is:  30 units long