Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ball’s journey will gravity be the greatest force acting on the ball?

Answers

Answer 1

Answer:

If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.


Related Questions

A 10,000J battery is depleted in 2h. What power consumption is this? *

A) 5000W
B) 3W
C) 1.4W
D) 20000W

show your work please

Answers

Answer:

P = 1.4 W

Explanation:

Given that,

The work done or the energy of the battery, E = 10,000 J

Time, t = 2 h

We need to find the power consumption. Let it is P. Power is the rate of doing work. So,

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W[/tex]

or

P = 1.4 W

So, the power of the battery is 1.4 W.

Why did vygotsky believe that children were more dependent learners

Answers

Answer:

Vygotsky's sociocultural theory asserts that learning is an essentially social process in which the support of parents, caregivers, peers and the wider society and culture plays a crucial role in the development of higher psychological functions.

Thus,  Vygotsky believed that children were more dependent learners

Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passes over the pulley as shown. Masses M1 and M3 lies on a 30o incline plane which slides down the plane. The coefficient of kinetic friction on the incline plane is 0.28. A. Draw a free body diagram of all the forces acting in the masses M1 and M2. B. Determine the tension in the string that connects M2 and M3.

Answers

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ  T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

T₁ = m₂aᵧ + m₂gT₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

∑Fₓ = maₓ F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

(F_n · μ_k) - m₁g sinΘ = m₁aₓ (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ [2.539595871] - [-58.0962595] = 6aₓ 60.63585537 = 6aₓ aₓ = 10.1059759 m/s²

Now let's go back to this equation:

T₁ = m₂(a + g)  

We have 3 known variables and we can solve for the tension force.

T = 2(10.1059759 + 9.8)T = 2(19.9059759)T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.

Answers

This question is incomplete, the missing diagram is uploaded along this answer below.

Answer:

- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

- the magnitude of compression force at the knee joint is 900 N

Explanation:

Given the data in the question and diagram below;

Net torque = 0

Torque = force × lever arm

so

F[tex]_{ConF[/tex]  × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

given that F[tex]_{ConF[/tex] = 90 N

90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in

90 N × 16.5 in =  T[tex]_{HonL[/tex] × 1.5 in

T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in

T[tex]_{HonL[/tex] = 990 N

Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N

b) magnitude of compression force at the knee joint;

In equilibrium, net force = 0

along horizontal

F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0

we substitute

F[tex]_{FonB[/tex] - 990 + 90 = 0

F[tex]_{FonB[/tex] - 900 = 0

F[tex]_{FonB[/tex] = 900 N

Therefore, the magnitude of compression force at the knee joint is 900 N

occurs when an air mass and its clouds encounter a mountain. This forces the air mass to move from a low elevation to a high elevation as it crosses over the mountain.

Frontal wedging
Orographic lifting
Localized convective lifting
Convergence
Jet streams

Answers

Answer:

Localized convective lifting

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Answer:

Explanation:

From the given information:

the car's momentum = momentum of the truck

(a) 816 kg × v = 2650 kg × 16.0 km/h

v = (2650 kg × 16.0 km/h) /  816 kg

v = 51.96 km/hr

(b) 816 kg × v = 9080 kg × 16.0 km/h

v = (9080 kg × 16.0 km/h) /  816 kg

v = 178.04 km/hr

what do you understand by Force,acceleration and mass??​

Answers

It states that the rate of change of velocity of an object is directly proportional to the force applied and takes place in the direction of the force. It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s²). Thus, an object of constant mass accelerates in proportion to the force applied.

Answer:

Force is a push or pull that an object can exert on other objects. Acceleration is the rate of change of an object's speed and Mass is the amount of matter in an object and is expressed in kilograms.

Explanation:

It is summarized by the equation: Force (N) = mass (kg) × acceleration (m/s2).

How do solar panels work with conduction, convection and radiation?

Answers

Answer:

In the case of a solar thermal panel we are trying to heat above the ambient temperature so conduction and convection will work against us by taking heat from the panel to the out- side world. ... The sun (at 6000 C surface temperature) is hotter than the solar panel so the panel will get hot due to the solar radiation.

Explanation:

Effective sex education must engage _____ more than _____.

Answers

Answer:

pregnant

Explanation:

no interest at school

To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direction of rain...​

Answers

Answer:

The actual speed of the rain is 5 m/s and its direction is -53.13°

Explanation:

The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.

V = v + v'

We add these vectorially.

Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i  and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j

So, V = v + v'

V = 3i + (-4j)

V = 3i - 4j

So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s

The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°

So, the actual speed of the rain is 5 m/s and its direction is -53.13°

How are hypotheses tested?

Answers

Answer:

by making observation hope it's helpful

By making an observation :)

How many molecules do we have for Na2Co3?

Answers

105.9888 g/mol is the mass as far as i know, Don't know the amount of molecules tho.

mark me brainliestt :))

The basal metabolic rate is the rate at which energy is produced in the body when a person is at rest. A 157 lb (71.0 kg ) person of height 5.91 ft(1.80 m ) would have a body surface area of approximately 1.90 m2 .

Reqiuired:
a. What is the net amount of heat this person could radiate per second into a room at 19.0 ∘C (about 66.2∘F) if his skin's surface temperature is 31.0 ∘C? (At such temperatures, nearly all the heat is infrared radiation, for which the body's emissivity is 1.00, regardless of the amount of pigment.)

b. Normally, 80.0 % of the energy produced by metabolism goes into heat, while the rest goes into things like pumping blood and repairing cells. Also normally, a person at rest can get rid of this excess heat just through radiation. Use your answer to part A to find this person's basal metabolic rate(BMR).

Answers

Answer:

A. Net amount of heat radiated = 109.2W

B. Person's basal energy = 136.5

Explanation:

Part A:

Area of person, A = 1.90 m^2

Temperature of person , T = 31 C

T = 304 K

Temperature of surroundings , To = 19 C

To = 282 K

Now, net amount of heat radiated = e*A*sigma *(T^4 - To^4)

Net amount of heat radiated = 1 * 1.8 * 5.6703 *10^-8 *(304^4 - 294^4)

Net amount of heat radiated = 109.2 W

The net amount of heat radiated is 109.2 W

Part B:

Person's basal energy = net amount of heat radiated /(0.80)

Person's basal energy = 109.2/0.80

Person's basal energy = 136.5 W

Person's basal energy is 136.5 W

calculate the electric potential 3mm from a point charge of 16Nc​

Answers

[tex]4.8 \times 10^8[/tex] volts

Explanation:

The electric potential due to a point charge is given by

[tex]V= \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q}{r}[/tex]

where Q = charge = [tex]16 \times 10^{-9}[/tex] C

r = distance from a point = [tex]3 \times 10^{-3}[/tex] m

[tex]\varepsilon_{0}[/tex] = permitivity of free space

= 8.85×10^-12 C^2/N-m^2

Plugging in the numbers,

[tex]V = \dfrac{1}{4 \pi (8.85 \times 10^{-12})} \dfrac{16 \times 10{-9}}{3 \times 10^{-3}}[/tex]

[tex]= 4.8 \times 10^8[/tex] volts

A person is driving a car down a straight road. The instantaneous acceleration is constant and in the direction of the car's motion. 1) The speed of the car is increasing. decreasing. constant. increasing but will eventually decrease. decreasing but will eventually increase.

Answers

Answer:Increasing

Explanation:

Given

Car is driven on the straight road with  instantaneous acceleration in the direction of car's motion.

If instanateneous acceleration is constant then speed of car is increasing at a constant pace. As there are no turns on the road, therefore speed of car is increasing.

The speed of the car is "decreasing". A further description is provided in the below paragraph.

It's because the individual would be in a straightforward fashion. This same acceleration inclination comes contrary to the movement of the automobile. It indicates that it exerts pressure against the movement of the automobile. So, when it moves forward, the speed of the automobile decreases.

Thus the above answer is correct.

Learn more about the speed here:

https://brainly.com/question/5053192

The magnification produced by spherical mirror is + 1/4. State the type of spherical mirror. State 3 characteristics of the image formed by the mirror: -​

Answers

Answer:

Convex mirror.

Explanation:

Image is real.

Image is inverted.

Image is magnified.

A cell phone weighs about 28x10" pounds. Which value of n is most reasonable?

Answers

Answer:

[tex] {3}^{n} [/tex]

Explanation:

https://www.doubtnut.com/question-answers/a-cellphone-weighs-about-28x10n-pounds-which-value-of-n-is-most-reasonable-a-3-b-2-c-0-d-1-433477

A charged particle accelerates as it moves from location A to location B. If VA = 260 V and VB = 210 V, what is the sign of the charged particle? positive negative (b) A electron loses electric potential energy as it moves from point 1 to point 2. Which of the following is true regarding the electric potential at points 1 and 2?

Answers

Answer:

(a) Positive

(b) Electron gains energy as it moves from A to B.

Explanation:

VA = 260 V

VB = 210 V

An electron moves from lower to higher potential which is negatively charged and a positively charged particle moves  from higher to lower potential, so the charge particle is positive in nature.

(a) Positive

(b) No, electron gains energy as it moves from A to B.

Can someone please help me with this problem

Answers

Answer:

resultant is equal to the sum of A vector or B vector and draw resultant in order that the tail of resultant vector concides with tail of vector a and head of resultant concides with the head of vector b

Explanation:

I provided the question above.

Answers

Answer:

Explanation:

since it is connected in parallel combination

use this formula

[tex]\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2}[/tex]

[tex]\frac{1}{R} = \frac{1}{2} + \frac{1}{4}[/tex]

[tex]\frac{1}{R} = \frac{4+2}{2}[/tex]

[tex]\frac{1}{R} = \frac{6}{2}[/tex]

[tex]\frac{1}{R} = 3[/tex] ohm

therefore resistence = 3 ohm

then we should find power

P = VI

P = 12*3

P = 24 watt

now to find current use formula power = current * voltage

24 = current * 12

24/12 = current

2 = current

therefore current is 2 ampere (A).

to find potential difference (emf) use formula

V = IR

V = current * resistence

V = 2 * 3

V = 6 volt .

therefore potential difference is 6 volt.

Choose the CORRECT statements. The superposition of two waves.

I. refers to the effects of waves at great distances.

Il. refers to how displacements of the two waves add together.

Ill. results into constructive interference and destructive interference

IV. results into minimum amplitude when crest meets trough.

V. results into destructive interference and the waves stop propagating.

A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V​

Answers

Answer:

A

Explanation:

I guess not that much confidential!

A light hollow tube of 2.00 cm diameter and 1.0 m length is filled with tiny beads of different density. The resulting density distribution is linear, with the left end having a density of 1.6 g/cm^3, and the right end having a density of 6.3 g/cm^3. How far from the left end will be the center of mass? (give answer in cm).

Answers

Answer:

i think c

Explanation:

  cause

What is the internal resistance of a current source with an EMF of 12 V if, when a resistor with an unknown resistance is connected to it, a current of 2 A flows through the circuit? A voltmeter connected to the source terminals shows 8 V.

Answers

Explanation:
Second question:
U
=
I
2
×
R
2
=
4
A
×
9
Ω
=
36
V
First question:
You could use
1
R
=
1
R
1
+
1
R
2
+
1
R
3
=
1
18
+
1
9
+
1
6
=
1
3

R
=
3
Ω
Or:
You can calculate the currents through the other resistors, add them all up and recalculate the total resistance (voltage already calculated):
I
1
=
U
R
1
=
36
V
18
Ω
=
2
A
I
2
=
4
A
(as given)
I
3
=
U
R
3
=
36
V
6
Ω
=
6
A
Now
R
=
U
I
=
36
V
2
A
+
4
A
+
6
A
=
36
V
12
A
=
3
Ω

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:

Answers

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = [tex]\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}[/tex]

  = [tex]4.59\times 10^{-19} \ J[/tex]

or,

  = [tex]\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}[/tex]

  = [tex]2.87 \ eV[/tex]

(b)

As we know,

⇒ [tex]Vq=\frac{hc}{\lambda}-\Phi_0[/tex]

By substituting the values, we get

⇒ [tex]1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0[/tex]

⇒                       [tex]\Phi_0=2.3\times 10^{-19} \ J[/tex]

or,

⇒                            [tex]=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}[/tex]

⇒                            [tex]=1.4375 \ eV[/tex]

When you are standing without moving, you exert a force on the ground. Why doesn't Earth slowly start accelerating downwards? ​

Answers

Explanation:

You would think it should. But remember the Force is also determined by mass. The mass of the earth markes our mass like the smallest part of a mosquito leg. The earth will go on it its merry way without cosidering us at all.

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Answer:

(a) v₁ = 51.96 km/h

(b) v₁ = 178 km/h

Explanation:

(a)

For having the same momentum:

m₁v₁ = m₂v₂

where,

m₁ = mass of Volkswagen = 816 kg

v₁ = speed of Volkswagen = ?

m₂ = mass of Cadillac = 2650 kg

v₂ = speed of Cadillac = 16 km/h

Therefore, using these values in the equation, we get:

[tex](816\ kg)v_1 = (2650\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{2650\ kg}{816\ kg})[/tex]

v₁ = 51.96 km/h

(b)

For having the same momentum:

m₁v₁ = m₂v₂

where,

m₁ = mass of Volkswagen = 816 kg

v₁ = speed of Volkswagen = ?

m₂ = mass of Truck = 9080 kg

v₂ = speed of Truck = 16 km/h

Therefore, using these values in the equation, we get:

[tex](816\ kg)v_1 = (9080\ kg)(16\ km/h)\\\\v_1 = (16\ km/h)(\frac{9080\ kg}{816\ kg})[/tex]

v₁ = 178 km/h

How fast much an 816kg Volkswagen travel to have the same momentum as (a) a 2650kg Cadillac going 16.0 km/h? (b) a 9080-kg truck also going 16.0 km/hr?

Answers

Explanation:

(a) The momentum of a 2650kg Cadillac going 16.0 km/h or 4.44 m/s is :

p = 2650 kg × 4.44 m/s

= 11766 kg-m/s

(b) The momentum of a 9080-kg truck also going 16.0 km/hr or 4.44 m/s is :

p =mv

= 9080-kg × 4.44 m/s

= 40315.2 kg-m/s

If p = 11766 kg-m/s, velocity is :

[tex]v=\dfrac{p}{m}\\\\v=\dfrac{11766 }{816}\\\\=14.41\ m/s[/tex]

If p = 40315.2 kg-m/s, velcocity is :

[tex]v=\dfrac{p}{m}\\\\v=\dfrac{40315.2}{816}\\\\=49.40\ m/s[/tex]

Hence, this is the required solution.

A 500 Kg block is attached with a rope of length 5m, having area 0.4× 10-4 m2. If final length is 6m, Calculate the Stress, Strain and Young's Modulus?

Answers

Answer:

stress = 1.225 x 10^8 N/m^2

strain = 1/5

Young's modulus = 6.125 x 10^8 N/m^2

Explanation:

mass, m = 500 kg

length, L = 5 m

Area, A = 0.4 x 10^-4 m^2

Final length, L' = 6 m

extension, x = L'-L= 6 - 5 = 1 m

Stress is defined as force per unit area.

[tex]stress =\frac{Force}{Area}\\\\stress =\frac{500\times 9.8}{0.4\times 10^{-4}}\\\\stress = 1.225\times 10^8 N/m^2[/tex]

Strain is defined as the ratio of change in length to the original length.

[tex]strain =\frac{x}{L}\\\\strain = \frac{1}{5}[/tex]

Young's modulus is given by the ratio of stress to the strain.

[tex]Y = \frac{1.225 \times 10^8}{\frac{1}{5}}\\\\Y = 6.125\times 10^8 N/m^2[/tex]

It takes a minimum distance of 98.26 m to stop a car moving at 17.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Answers

Answer:

x_f = 212.5m

Explanation:

t = (x_f-x_0)/(.5*(v_f-v_0))

t = (98.26m-0m)/(.5(0m/s-17m/s))

t = 11.56s

a = (v_f-v_0)/t

a = (0m/s-17m/s)/11.56s

a = -1.47m/s²

t = (v_f-v_0)/a

t = (0m/s-25m/s)/-1.47m/s²

t = 17s

x_f = x_0+(.5*(v_f-v_0))*t

x_f = 0m+(.5*(0m/s-25m/s))*17s

x_f = 212.5m

what is the power of an electrical device which operates with a current of 12.4 A and a potential difference of 12 V​

Answers

148.8 Watts

Explanation:

P = VI

= (12 V)(12.4 A)

= 148.8 Watts

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