Can someone please help me with this?

Answer:

Therefore, the temperature at which the Fahrenheit scale reading is equal to half of the Celsius scale is −24.6∘C .

Answer:

The relation between velocity and time is a simple one during uniformly accelerated, straight-line motion. The longer the acceleration, the greater the change in velocity. Change in velocity is directly proportional to time when acceleration is constant.

~Hoped this helped~

~Brainiliest?~

Answer:

We describe motion in terms of velocity and acceleration. Velocity: The rate of change of displacement of an object (displacement over elapsed time) is velocity. Velocity is a vector since it has both magnitude (called speed) and direction. ... Acceleration: The rate of change of velocity is acceleration.

Explanation:

Displacement is a vector which points from the initial position of an object to its final position. ... Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity.

Displacement is the vector difference between the ending and starting positions of an object. Velocity is the rate at which displacement changes with time. ... The average velocity over some interval is the total displacement during that interval, divided by the time.

Hope this helps      :)

Answer:

The First is an example of acceleration, the second is an example of velocity

Answer:

left = 6:19 pm ; middle = 6:56 pm ; right = 9:05 am

Explanation:

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

Answer:

I think it's D

Explanation:

I don't know I just think

Answer:

θ ≈ 65°

Explanation:

From the question given above, the following data were obtained:

Maximum height (H) = 8 m

Range (R) = 15 m

Initial velocity (u) = v

Angle θ =?

H = u²Sine²θ / 2g

8 = v²Sine²θ / 2g

Cross multiply

8 × 2g = v²Sine²θ

16g = v²Sine²θ

Divide both side by Sine²θ

v² = 16g / Sine²θ.... (1)

R = u²Sine2θ / g

15 = v²Sine2θ / g

Cross multiply

15 × g = v²Sine2θ

15g = v²Sine2θ

Divide both side by Sine2θ

v² = 15g / Sine 2θ... (2)

Summary:

v² = 16g / Sine²θ.... (1)

v² = 15g / Sine 2θ... (2)

Equate equation 1 and 2

16g / Sine²θ = 15g / Sine 2θ

16 / Sine²θ = 15 / Sine 2θ

Recall:

Sine²θ = SineθSineθ

Sine 2θ = 2SineθCosθ

16 / Sine²θ = 15 / Sine 2θ

16 / SineθSineθ = 15 / 2SineθCosθ

16 / Sineθ = 15 / 2Cosθ

Cross multiply

15 × Sineθ = 16 × 2Cosθ

15Sineθ = 32Cosθ

Divide both side by Cosθ

15Sineθ / Cosθ = 32

Divide both side by 15

Sineθ / Cosθ = 32/15

Recall:

Sineθ / Cosθ = Tanθ

Sineθ / Cosθ = 32/15

Tanθ = 32/15

Tanθ = 2.1333

Take the inverse of Tan

θ = Tan¯¹ 2.1333

θ ≈ 65°

Answer:

Momentum is a vector quantity that has the same direction as the velocity of the object. The quantity of force multiplied by the time it is applied is called impulse. Impulse is a vector quantity that has the same direction as the force. Momentum and impulse have the same units: kg·m/s

Explanation:

Hope it helps!

If you dont mind can you please mark me as brainlest?

A switched-mode power supply (switching-mode power supply, switch-mode power supply, switched power supply, SMPS, or switcher) is an electronic power supply that incorporates a switching regulator to convert electrical power efficiently.

Answer:

A switched-mode power supply (switching-mode power supply, switch-mode power supply, switched power supply, SMPS, or switcher) is an electronic power supply that incorporates a switching regulator to convert electrical power efficiently.

Explanation:

Answer:

Blue car

Explanation:

first let's find the speed:

1) S=D/T

so 120/1= 120km per hr

2) 130/1=130km per hr

so as we can see 130km per hr is faster than 120 km per hr by 10 km per hr

Answer:

papa's donuteria

Explanation:

:rawr:

Answer; Sodium 2-Aminopentanedioate The chemical formula of MSG is C5H8NO4Na and its IUPAC name is sodium 2-aminopentanedioate. Since it is known to intensify meaty flavours in food, monosodium glutamate is widely used as a flavour enhancer in the food industry. This compound was first produced by the Japanese chemist Kikunae Ikeda in the year 1908.

Explanation:

Answer:the answer is conduction

Answer: Use the observations made in the virtual lab to complete this table. (pg. 5)

Month in

General observations of surface temperature anomaly maps

2008

Land

Sea

Hide side

Explanation: Use the observations made in the virtual lab to complete this table. (pg. 5)

Month in

General observations of surface temperature anomaly maps

2008

Land

Sea

Hide side

Answer:

The electric force decrease as the distance increase and vice versa

As it obeys inverse square law that states ththaforce is inversly proportional to square if distance

so if the distance increased to twice the force will be decreased to quarter.

And that is coloumb's law of electrostatic force:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

F=k(Q1.Q2)/r2

So as you increase the distance between charged objects, Force(attraction/repulsion) weakens by the square of their distance.

As you decrease the distance between charged objects, Force strengthens by the square of their distance.

Inverse relationships are common in nature. In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance. These observations are commonly made during demonstrations and lab experiments. Consider a charged plastic golf tube being brought near a collection of paper bits at rest upon a table. The electrical interaction is so small at large distances that the golf tube does not seem to exert an influence upon the paper bits. Yet if the tube is brought closer, an attractive interaction is observed and the strength is so significant that the paper bits are lifted off the table. In a similar manner, charged balloons are observed to exert their greatest influence upon other charged objects when the separation distance is reduced. Electrostatic force and distance are inversely related.

The pattern between electrostatic force and distance can be further characterized as an inverse square relationship. Careful observations show that the electrostatic force between two point charges varies inversely with the square of the distance of separation between the two charges. That is, the factor by which the electrostatic force is changed is the inverse of the square of the factor by which the separation distance is changed. So if the separation distance is doubled (increased by a factor of 2), then the electrostatic force is decreased by a factor of four (2 raised to the second power). And if the separation distance is tripled (increased by a factor of 3), then the electrostatic force is decreased by a factor of nine (3 raised to the second power). This square effect makes distance of double importance in its impact upon electrostatic

Answer:

Charge

Explanation:

Answer:

1.55

Explanation:

that it is the standard refractive index.

Answer:

The answer is D for plato users :)

Explanation:

Answer:

the answer is all of the above

Explanation:

brainliest please

Answer:

2m/s^2

Explanation:

Force is equal to;

mass × acceleration

therefore;

acceleration equals

force divided by mass

acceleration = 2m/s ^2

Answer:

f = 60N

m = 30kg

f= m×a

60N= 30 × a

a = 60/30

a = 2

Answer:

Follows are the soplution to this question:

Explanation:

In the given scenario it would be like a fluid in a simple harmonic in contact with the earth. Whenever a cheetah hops quicker, oscillatory amplitude rises, while the duration stays the same since it does not depend on frequency, which mostly means that time will be the same if you're in contact with the substrate.

Answer:

yellow

Explanation:

yellow

make me brain

Answer:

Assume that [tex]g =9.81\; \rm m\cdot s^{-1}[/tex], and that the air resistance on the stone is negligible.

a.

Height of the stone: [tex]389.5\; \rm m[/tex] (above the ground.)

Velocity of the stone: [tex]\left(-110.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)

b.

Height of the stone: [tex]629.5\; \rm m[/tex] (above the ground.)

Velocity of the stone: [tex]\left(-86.5\; \rm m \cdot s^{-1}\right)[/tex] (the stone is travelling downwards.)

Explanation:

If air resistance on the stone is negligible, the stone would be accelerating downwards at a constant [tex]a = -g = -9.81\; \rm m \cdot s^{-2}[/tex].

Let [tex]h_0[/tex] denote the initial height of the stone (height of the stone at [tex]t = 0[/tex].)

Similarly, let [tex]v_0[/tex] denote the initial velocity of the stone.

Before the stone reaches the ground, the height [tex]h[/tex] (in meters) of the stone at time [tex]t[/tex] (in seconds) would be:

[tex]\displaystyle h(t) = -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0[/tex].

Similarly, before the stone reaches the ground, the velocity [tex]v[/tex] (in meters-per-second) of the stone at time [tex]t[/tex] (in seconds) would be:

[tex]v(t) = -g\cdot t + v_0[/tex].

In section a., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = -12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling downwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:

[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + \left(-12\; \rm m \cdot s^{-1}\right) \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 389.5\; \rm m \end{aligned}[/tex].

Indeed, the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is greater than zero. The stone hasn't yet hit the ground, and both the representation for the height of the stone and that for the velocity of the stone are valid.

[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s - 12\; \rm m\cdot s^{-1} \\ &= -110.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].

The value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative, meaning that the stone would be travelling downwards at that time.

In section b., [tex]h_0 = 1000\; \rm m[/tex] while [tex]v_0 = 12\; \rm m\cdot s^{-1}[/tex] (the stone is initially travelling upwards.) Evaluate both [tex]h(t)[/tex] and [tex]v(t)[/tex] for [tex]t = 10\; \rm m \cdot s^{-1}[/tex]:

[tex]\begin{aligned} h(t) &= -\frac{1}{2}\, g \cdot t^{2} + v_0 \cdot t + h_0 \\ &= -\frac{1}{2}\ \times 9.81\; \rm m\cdot s^{-2}\times (10\; \rm s)^{2} \\&\quad\quad + 12\; \rm m \cdot s^{-1} \times 10\; \rm s + 1000\; \rm m \\[0.5em] &= 629.5\; \rm m \end{aligned}[/tex].

Verify that the value of [tex]h(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is indeed greater than zero.

[tex]\begin{aligned} v(t) &= -g\cdot t + v_0 \\ &= -9.81\; \rm m\cdot s^{-2}\times 10\; \rm s + 12\; \rm m\cdot s^{-1} \\ &= -86.5\; \rm m \cdot s^{-1} \end{aligned}[/tex].

Similarly, the value of [tex]v(t)[/tex] at [tex]t = 10\; \rm m \cdot s^{-1}[/tex] is negative because the stone would be travelling downwards at that time.

[tex]\lambda=\frac{v}{f}=\frac{400}{200}=2[m][/tex]

Answer:

sound waves

Explanation:

this is because sound waves are longitudinal waves, and longitudinal waves are waves that travel parallel to the direction of the wave motion

thus it cannot be light or electromagnetic waves but only sound waves

hope this helps, please mark it

Answer:

G

Explanation:

That is is the equator, prolly

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

Answer:

the force of gravity acting between the earth and any other object is directly proportional to the mass of the earth, directly proportional to the mass of the object, and inversely proportional to the square of the distance that separates the centers of the earth and the object.

Explanation: