Convert this number to standard notation: 1.4 x 10^-2*

Answer:

D because those are both concerning.

Answer:

the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Explanation:

Given that;

final velocity v = 0

initial velocity u = 15m/s

time taken t = 4 s

acceleration  a = ?

from the equation of motion        

v   =   u   +   at

we substitute

0 = 15 + a × 4

acceleration a = -15/4 =  - 3.75 m/s²    

the negative sign tells us that its a  deacceleration so the sign can be ignored.

Deacceleration due to friction a = μ × g

we substitute

3.75 = μ × 9.8    

μ = 3.75 / 9.8 = 0.3826 ≈ 0.38

Therefore the coefficient of Kinetic friction between the tires and road is 0.38

Option A) .38 is the correct answer

Write the velocity vectors in component form.

• initial velocity:

v₁ = 4 m/s at 45º N of E

v₁ = (4 m/s) (cos(45º) i + sin(45º) j)

v₁ ≈ (2.83 m/s) i + (2.83 m/s) j

• final velocity:

v₂ = 4 m/s at 10º N of E

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ ≈ (3.94 m/s) i + (0.695 m/s) j

The average acceleration over this 3-second interval is then

a = (v₂ - v₁) / (3 s)

a ≈ (0.370 m/s²) + (-0.711 m/s²)

with magnitude

||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²

and direction θ such that

tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92

→   θ ≈ -62.5º

which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.

Answer:

Good question to ask in physics, sir maam

Answer:

That's good

Explanation:

Answer:

[tex]F'=4.5\times 10^{-5}\ N[/tex]

Explanation:

The gravitational force between two pickups is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

It means,

[tex]F\propto m_1m_2[/tex]

So,

[tex]\dfrac{F}{F'}=\dfrac{m_1m_2}{m_1'm_2'}[/tex]

We have,

[tex]m_1=m_2=2000\ kg\\\\F=3\times 10^{-5}\ N\\\\m_1'=2000+1000 =3000\ kg\\\\m_2'\ \text{remains the same i.e. 1000 kg}[/tex]

F' is the new force

So, putting all the values,

[tex]\dfrac{3\times 10^{-5}}{F'}=\dfrac{2000\times 2000}{3000\times 2000}\\\\\dfrac{3\times 10^{-5}}{F'}=\dfrac{2}{3}\\\\F'=\dfrac{9\times 10^{-5}}{2}\\\\F'=4.5\times 10^{-5}\ N[/tex]

So, the new force between the two trucks is [tex]4.5\times 10^{-5}\ N[/tex].

Answer:

V₁ = 5 m/s

This is the problem of one dimension

Explanation:

Before using the concepts of relative velocity, we must first define the variables for each velocity here:

V₁ = Person's Velocity Relative to Pier = ?

V₂ = Boat's Velocity Relative to Pier = 6 m/s

V₁₂ = Person's Velocity Relative to Boat = - 1 m/s

Now, from the relative velocity, we know that:

V₁₂ = V₁ - V₂

- 1 m/s = V₁ - 6 m/s

V₁ = 6 m/s - 1 m/s

V₁ = 5 m/s

This is the problem of one dimension

because, the velocity is along single dimension with opposite directions. No, second dimension is involved.

Answer:

Workdone = 25500Nm

Explanation:

Given the following data

Force = 425N

Distance = 60m

To find the workdone

Workdone = force *distance

Substituting into the equation, we have

Workdone = 425*60

Workdone = 25500Nm

Answer:

I've attached a screenshot of a document from K12 that should give you the answer to your question, and many more question you might have. If this is not helpful please say so in the replies to this answer and I'll try my best to find some more information.

Explanation:

May I have brainliest please? :)

Answer:

d. Direction and magnitude

Explanation:

The two components of a vector are its magnitude and direction.

Magnitude is the quantity of the substance

Direction is the path.

Examples of vector quantities are velocity, displacement, acceleration.

Answer:

A and B.

Explanation:

Answer C is wrong - there are no tall trees underwater

Answer D is wrong, there are mountain chains underwater due to plate tectonics.

Answer:

Many drivers follow the “three-second rule.” In other words, you should keep three seconds worth of space between your car and the car in front of you in order to maintain a safe following distance.

Thank you and please rate me as brainliest as it will help me to level up

Answer:

(A) q/m = 9.58 x 10⁷ C/kg

(B) b. Proton

Explanation:

(A)

In order to pass un-deflected between magnetic and electric field both electric and magnetic forces must be equal:

Felectric = Fmagnetic

Eq = Bqv

v = E/B   ------------ equation (1)

Now, when the particle forms semi-circle under influence of magnetic field. At this, point magnetic field is equal to centripetal force:

Centripetal Force = Magnetic Force

mv²/r = qvB

mv = qrB

v = qrB/m   ------------ equation (2)

comparing equation (1) and equation (2), we get:

E/B = qrB/m

q/m = E/B²r

where,

q/m = charge to mass ratio = ?

E = Electric Field = 187500 V/m

B = Magnetic Field = 0.125 T

r = radius of circular path = 25.05 cm/2 = 12.525 cm = 0.12525 m

Therefore,

q/m = (187500 V/m)/(0.125 T)²(0.12525 m)

q/m = 9.58 x 10⁷ C/kg

(B)

This is the charge to mass ration of a proton:

q/m = 1.6 x 10⁻¹⁹ C/1.67 x 10⁻²⁷ kg

q/m =  9.58 x 10⁷ C/kg

Therefore, correct option is:

b. proton

Answer:

Approximately [tex]4.39 \; \rm m \cdot s^{-1}[/tex] (assuming that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)

Explanation:

Let the initial velocity of this stone be [tex]v\; \rm m \cdot s^{-1}[/tex] ([tex]v \ge 0[/tex] because speed should be non-negative.) That should be numerically equal to the initial horizontal speed of this stone. Because the stone was thrown horizontally, its vertical speed would initially be zero ([tex]0\; \rm m \cdot s^{-1}[/tex].)

The weight of this stone will accelerate the stone downwards. Because the weight of the stone is in the vertical direction, it will have no effect on the horizontal speed of the stone.

Therefore, if air resistance is indeed negligible, the horizontal speed of this stone will stay the same. The horizontal speed of the stone after [tex]0.5\; \rm s[/tex] should still be [tex]v\; \rm m \cdot s^{-1}[/tex]- same as its initial value.

On the other hand, if [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the (downward) vertical speed of this stone will increase by [tex]9.81\; \rm m \cdot s^{-2}[/tex] every second. After [tex]0.5\; \rm s[/tex], the vertical speed of this stone would have become:

[tex]0.5\; \rm s \times 9.81\; \rm m \cdot s^{-2} = 4.905\; \rm m \cdot s^{-1}[/tex].

[tex]t = 0\; \rm s[/tex]:

[tex]t = 0.5\; \rm s[/tex]:

Refer to the diagram attached. Consider the vertical speed and the horizontal speed of this rock as lengths of the two legs of a right triangle. The numerical value of the velocity of this stone would correspond to the length of the hypotenuse of that right triangle. Apply the Pythagorean Theorem to find that numerical value:

[tex]\begin{aligned} &\text{numerical value of velocity} \\ &= \sqrt{{\left(\text{horizontal speed}\right)}^2 + {\left(\text{vertical speed}\right)}^2}\end{aligned}[/tex].

At [tex]t = 0\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\sqrt{v^{2} + 0^2} = v\; \rm m \cdot s^{-1}[/tex].

At [tex]t = 0.5\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\left(\sqrt{v^2 + 4.905^2}\right)\; \rm m \cdot s^{-1}[/tex].

The question requires that the magnitude of the velocity of the stone at [tex]t = 0.5\; \rm s[/tex] should be [tex]1.5[/tex] times the value at [tex]t = 0\; \rm s[/tex]. In other words:

[tex]\sqrt{v^2 + 4.905^2} = 1.5 \, v[/tex].

Square both sides of this equation and solve for [tex]v[/tex]:

[tex]v^2 + 4.905^2 = 2.25\, v^2[/tex].

[tex]v \approx 4.39[/tex] ([tex]v \ge 0[/tex] given that speed should be non-negative.)

Therefore, the initial velocity of this stone should be approximately [tex]4.39\; \rm m \cdot s^{-1}[/tex].

Explanation:

That must be the right ans.

Answer:

I think it's c .... fulcrum

Answer:

F = 7,916,955.0N

Explanation:

According to newtons second law

Force = mass * acceleration

Given

mass = 52.0kg

distance S = 22.0m

time t = 17.0 ms = 0.017s

We need to get the acceleration first using the formula;

S = ut+ 1/2at²

22 = 0 + 1/2 a(0.017²)

22 = 0.0001445a

a = 22/0.0001445

a = 152,249.13m/s²

The magnitude of the average force exerted will be;

F = ma

F = 52 * 152,249.13

F = 7,916,955.0N

Answer:

40N

160N

360N

Explanation:

The drag force can be calculated using the formula below

FD = 0.5CPAV²

The drag coefficient = c = 0.8

The frontal density = p = 1.0kg

The frontal surface area = A = 1.0

The velocity v = 10m/s, 20m/s, 30m/s

When V = 10m/s

Fd = 0.5(0.8)(1)(1)(10)²

Fd = 40N

When V = 20m/s

FD = 0.5(0.8)(1)(1)(20)²

FD = 160N

When V = 30m/s

FD = 0.5(0.8)(1)(1)(30)²

FD = 360N

Thank you

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

A)The magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor will be 26.58 Nm.

Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.

The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.

m is the mass of steel ball = 3.0 kg in his hand.

L is the length of the arm is 60 cm long

M is the mass of arm=3.8 kg,

The torque is given as;

The magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor is found as;

[tex]\rm \tau_1 = F \times d \\\\ \rm \tau_1 = 29.4 \times 0.63 \\\\ \rm \tau_1 =17.64 \ Nm[/tex]

The torque due to the arm;

[tex]\rm \tau_2= F \times d \\\\ \rm \tau_2 = mg \times d \\\\ \rm \tau_2 =3.8 \times 9.81 \times 0.24 \ Nm \\\\ \rm \tau_2=8.94 \ Nm[/tex]

The net torque for case 1 will be;

[tex]\tau = \tau_1 + \tau_2\\\\ \rm \tau = 17.64 Nm + 8.94 Nm\\\\ \tau = 26.58 Nm[/tex]

Hence the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor will be 26.58 Nm.

B) the magnitude of the torque about his shoulder if he holds his arm straight, but 45∘ below horizontal will be 18.79 Nm.

The torque is due to the weight of the ball;

[tex]\rm \tau_1= F \times d \\\\ \rm \tau_1 = mg \times d \\\\ \rm \tau_1 =3.8 \times 9.81 \times 0.4242 \ Nm \\\\ \rm \tau_1= 12.47 \ Nm[/tex]

The torque due to the arm will be;

[tex]\rm \tau_2= F \times d \\\\ \rm \tau_2 = mg \times d \\\\ \rm \tau_2 =3.8 \times 9.81 \times 0.1696 \ Nm \\\\ \rm \tau_2=6.32 \ Nm[/tex]

The net torque in case 2 will be

[tex]\tau = \tau_1 + \tau_2\\\\ \rm \tau = 12.47 Nm + 6.32 Nm\\\\ \tau = 18.79 Nm[/tex]

Hence the magnitude of the torque about his shoulder if he holds his arm straight, but 45° below horizontal will be 18.79 Nm

To learn more about the torque refer to the link;

brainly.com/question/6855614

Answer:

The Answer is (A) . . .

I Think . . .

Answer:

a. buoyancy

....................

Answer:

The gravity arrow for each body rotates, always pointing toward the other body. Both arrows grow longer when the bodies come closer to one another and shorter when they move farther apart. This change shows that the gravitational force is stronger the closer together the bodies are.

Explanation:

A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies. Thus, a gravitational field—measured in newtons per kilogram (N/kg)—is employed to describe gravitational processes.

Each body's gravity arrow spins while constantly pointing in the direction of the other body. When the bodies move apart, the arrows go shorter and get longer as they get closer together. This modification demonstrates that the gravitational force increases with increasing body proximity.

Therefore, All objects with mass are attracted to one another by the gravitational attraction. Which has a magnitude that is directly proportional to the masses of the two objects and inversely proportional to the square of their distance from one another.

Learn more about gravitational force here:

https://brainly.com/question/12528243

#SPJ2

Answer:

The pectoralis major, latissimus dorsi, deltoid, and rotator cuff muscles connect to the humerus and move the arm. The muscles that move the forearm are located along the humerus, which include the triceps brachii, biceps brachii, brachialis, and brachioradialis.

Answer:

people who had a hard time to help other people because they feel that if they help

that the work of the person is much harder than his own work

Individuals who find it difficult to help others because they believe that if they do, the person's job will be much tougher than their own.

A group of people working together to achieve a shared purpose.

Collaboration, cooperation, and coordination are all words that come to mind when thinking of teaming.

Learn more:

https://brainly.com/question/10552710?referrer=searchResults

Answer:

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

Answer:

The correct option is B

Explanation:

Active galaxies are characterized by a small core of emission at the center which is usually brighter than the rest of the galaxy. The energy emitted from active galaxy is way more than that emitted from normal galaxy. As compared to normal galaxy where it's energy emission is thought to be the sum of the emission from each star found in the galaxy, this is not true or the same in active galaxy. There (active galaxy) huge energy emission cannot be explained as the sum of the emission of their stars.

Answer:

Explanation:

No doubt your curriculum materials have an adequate discussion of error and how it is determined.

a) The error between two values is the difference between an approximation and the true value. You determine it by subtracting the true value from the approximation.

In the case of a trip, you need to decide what is the "true value" and what is the "approximation." The wording here suggests that the "displacement of the actual trip taken" is to be considered the "true value." Then, you determine the error by subtracting that displacement from the "vector addition."

__

b) In Euclidean geometry, a vector is a straight-line path. In the real world, we use the term vector to refer to the direction taken by an object, generally along a curved path at some relative distance from the Earth's surface. The direction reference may change along the path, so that following a given "vector" will usually result in a curved path (not a great circle) on the Earth's surface.

__

c) The curved path from a point A to a point B on the Earth's surface will always be longer than the straight-line path between the same two points. In the case of points on opposite sides of the world, the straight-line path is through the center of the Earth, whereas the "vector addition" will be some path along the surface of the Earth.

You need to decide the meaning of "actual trip taken," as any actual trip will generally involve all the changes in direction and ups and downs along the way. If you consider "the actual trip taken" to be the difference between starting and ending coordinates, then the "vector addition" will always be longer.

Answer:

56.7kpa

Explanation:

V = 0.0138-m³

M = 4.73g

T = 358k

R = ideal gas constant = 8.314

Number of moles of water n = 4.73/18 = 0.2628 (Remember that the molecular mass of water is 18).

Pv = nRT

P = nRT/V

P = (0.2628x8.314x358)/0.0138

P = 56681.23

P = 56.7Kpa

56.7Kpa is therefore the pressure of the water vapor.