A golf ball is sitting on a tee. The ball is struck with a golf club and flies
through the air. How does the force on the club compare with the force on the
ball when momentum is transferred between the club and ball?

Answer:

my middle school life was horrible and still is sadly ); could you please do my recent brainless <333

Explanation:

My middle school life was eh.i allways got assumed that i was a elementry kid when i was and 6 and 7 grade.until 8th when i hit my grow spurt.my voice end up getting really deep in 8th grade too.tho in 8th i got sent to a certain  type of scholl for kids who did bad because of some dum teacher who lied that i hit her when i actually just went to her desk to get my phone sence she saw it in my pocket and took it away.i got sent there for 3 months and was worst thing ever.the treated us like prisoners in that school and made us tacky uniforms that i refused to where.i believe it was a school for 7th-12th.after my expierence at that school my expierence in 8th grade was almost done.i end up being popular at school for being that kid who hit the teacher.ofc that was like 2 years ago im 10th now.i dont really remember my 6th and 7th grade year.only thing i rember is i was a trouble maker and still am a trouble maker.

hope that helped you:)

please make my answer brainiest as im trying get 4 brainiest answers

Answer:

The mass of the object is 8.41 Kg.

Explanation:

Mechanical Force

The second Newton's law states that the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object. The SI unit for the force is the Newton: [tex]1\ Nw = 1~Kg.m/s^2[/tex]

We are given the net force of F=29.43 N acting on an object and producing an acceleration of a=3.5~m/s^2.

To calculate the mass of the object, we solve the above equation for m:

[tex]\displaystyle m=\frac{F}{a}[/tex]

[tex]\displaystyle m=\frac{29.43}{3.5}[/tex]

Calculating:

m = 8.41 Kg

The mass of the object is 8.41 Kg.

Answer:

i dont know dude but ask someone who is in your grade lol

Explanation:

Answer:

8649 J

Explanation:

KE = 1/2mv^2

1/2(50)(18.6)^2

1/2(50)(345.96) = 8649 J

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

[tex]F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r[/tex]

Where;

m = The mass of the person

v = The velocity of the person

[tex]F_c[/tex] = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for  the centripetal force, we have;

[tex]F_c[/tex] = m × ω² × r

1000 = 76.45 × (2·π/15)² × r

r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

Answer:                                                                              

Explanation:(40.0)×(9.8)=392

764N-392=372

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

Answer:

[tex]2.33\ m/s^2[/tex]

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn = m.a

Where a is the acceleration of the object.  The net force is the sum of the individual vector forces applied to the object.

The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.

The net force is Fn = 5000 N - 1500 N = 3500 N

To find the acceleration, we solve the equation for a:

[tex]\displaystyle a=\frac{Fn}{m}[/tex]

[tex]\displaystyle a=\frac{3500}{1500}[/tex]

[tex]\boxed{a = 2.33\ m/s^2}[/tex]

The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]

Answer:

a

Explanation:

because it can increase anything

Answer:

I think A

Explanation:

because it dosn't have enough tools

Answer:

3 hours 71 minutes

Explanation:

As because Speed= distance/time taken

so time taken= 3.71

Answer:

Tailed star

Explanation:

Tailed star

Answer:

8.5m/s

Explanation:

Using the equation of motion

v² = u² + 2gH

v is the final speed of the diver

u  is the initial speed of the diver

g is the acceleration due to gravity

H is the height of the object

Given the following

V = 1.3m/s

H = 3.60m

g = 9.8m/s²

Required

Initial speed of the diver u

Substitute the given values into the formula:

v² = u² -2gH

1.3² = u²  - 2(9.8)(3.60)

1.69 = u²-70.56

u² = 1.69+70.56

u² = 72.25

u = √72.25

u = 8.5m/s

Hence the diver's speed, in m/s, just before she enters the water is 8.5m/s

Answer:

450 g =0.9921 lb

Explanation:

It would take approximately 4 hours for the train to cover a distance of 470km.

4 hours 5 minutes 13.04 seconds

The acceleration at segment D is -20m/s²

The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)

Acceleration is the change in velocity with respect to time.

a = v-u/t

Acceleration at segment A:

Aa = 15-0/1-0

Aa = 15m/s²

Acceleration at segment B:

Ab = 20-15/2-1

Ab = 5m/s²

Acceleration at segment C:

Aa = 0-0/4-2

Aa = 0m/s²

Acceleration at segment D:

Ac = 0-20/5-4

Ac = -20m/s²

Hence the acceleration at segment D is -20m/s²

The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)

Learn more here: https://brainly.com/question/18398656

Answer:

He should swim at least three times a week at the community pool to build stamina.

Explanation:

Answer:

The final velocity, [tex]v_{f}[/tex], is gd.

Explanation:

The condition here is a free falling object. Thus from the third equation of motion under free fall, we have;

[tex]v_{f}[/tex] = [tex]v_{i}[/tex] + 2gs

where [tex]v_{f}[/tex] is the final velocity of the object, [tex]v_{i}[/tex] is the initial velocity of the object, g is the gravitational force and s is the height.

Since the object falls from a height of d, then [tex]v_{i}[/tex] = 0 m/s, and s = d.

So that;

[tex]v_{f}[/tex] = 0 + 2gd

   = 2gd

[tex]v_{f}[/tex] = 2gd

When the distance is [tex]\frac{d}{2}[/tex], [tex]v_{i}[/tex] = 0 m/s.

Then;

[tex]v_{f}[/tex] = 2g[tex]\frac{d}{2}[/tex]

[tex]v_{f}[/tex] = g x d

When the object falls through the height [tex]\frac{d}{2}[/tex], then the final velocity is gd.

Answer

Soap into a liquid

Explanation:

Given that,

Volume of gas, V₁ = 200 mL

Initial temperature, T₁ = 27°C

Final temperature, T₂ = -33°C

To find,

New volume of the gas.

Solution,

Initial temperature, T₁ = 27°C = 300 K

Final temperature, T₂ = -33°C = 240 K

Let V₂ is the new volume.

The relation between volume and temperature is given by :

[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\=\dfrac{200\times 240}{300}\\\\=160\ mL[/tex]

Hence, the volume will be 160 mL.

Answer:

The earth rotates west to east

Explanation:

Considering that the sun doesn't really move and the earth is rotating around the sun, and it is moving from east to west with us (the earth) facing it in which means the earth rotates around the sun west to east!

Answer:

the sun light will heat up the balloon causing it to burst

Answer:

iron

Explanation:

Answer:

Explanation:

Trypophobia is a fear or disgust of closely-packed holes. People who have it feel queasy when looking at surfaces that have small holes gathered close together.A person may start by imagining what they fear, then looking at pictures of the fear object, and then finally being near or even touching the source of their anxiety. In the case of trypophobia, a person with symptoms may start by simply closing his eyes and imagining something such as a honeycomb or seed poThey can help you find the root of the fear and manage your symptoms. Last medically reviewed on July 20, 2017 Medically reviewed by Timothy J. Legg, Ph.D., CRNP — Written by Annamarya Scaccia ...

Answer:

radius of a circle is measured by an instrument, so it is the independent variable

Explanation:

The independent variable corresponds to the magnitude controlled by the experimenter and in general can be measured by an instrument.

The dependent variable is the variable used to perform the calculations, obtained from a mathematical transformation of the independent variable.

The radius of a circle is measured by an instrument, so it is the independent variable and the area is calculated by

         A = π r²

Therefore the area corresponds to the dependent variable

Answer:

frictional force acting on the cart 450 N.

Given:

mass of cart = 50 kg

acceleration = 3

To find:

Frictional force = ?

Formula used:

F -  = 600 N

F = ma

= frictional force

Solution:

Force on the block is given by,

F -  = 600 N

F = ma

= frictional force

= 600 - m a

= 600 - (50 × 3)

= 600- 150

= 450 N

According to the given data the frictional force acting on the cart 450 N.

Answer:

B. [tex]5\,\frac{m}{s^{2}}[/tex] South

Explanation:

The correct statement is:

A washing machine is being dragged with a starting force of 380.8 N to the South, but this is reduced by friction. If the mass of the washing machine is 37 kg  and the coefficient of friction is 0.54, what was the Net acceleration experienced by the washing machine?

The net force of the washing machine moving on the horizontal ground is equal to the external force minus the friction force, where the normal force from the ground on the washing machine is equal to its weight. The free body diagram of the washing machine is presented below. The equation of equilibrium associated with the acceleration of the washing machine is:

[tex]\Sigma F = P -\mu_{k}\cdot m \cdot g = m\cdot a[/tex] (1)

Where:

[tex]P[/tex] - External force exerted on the washing machine, measured in newtons.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

[tex]m[/tex] - Mass, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]a[/tex] - Net acceleration of the washing machine.

If we know that [tex]P = 380.8\,N[/tex], [tex]m = 37\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\mu_{k} = 0.54[/tex], then the net acceleration experienced by the washing machine is:

[tex]a = \frac{P}{m}-\mu_{k}\cdot g[/tex]

[tex]a = \frac{380.8\,N}{37\,kg}-(0.54)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]a = 4.996\,\frac{m}{s^{2}}[/tex]

The correct answer is B.

Answer:

Explanation:

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