A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Temperature variations in a graph occur as a result of changing environmental conditions and changing temperature.  

what is temperature ?

Temperature is a physical quantity which measures hotness and coldness of a body. Temperature measures the degree of vibration of molecule in a body. Temperature is measured in centigrade (°C), Fahrenheit (°F) and Kelvin (K) in which Kelvin (K) is a SI unit of temperature. Absolute scale of temperature means Kelvin scale of temperature. relation between Kelvin(K) and centigrade (°C), °C= K - 273.15 from equation, 273.15 K means 0 °C, which is freezing point of water (ice). when we give temperature to the body, its molecule or atom absorbs thermal energy and vibrate about their mean position. Amplitude of vibration get increases as we go on increasing temperature and for higher temperature force of attraction between molecules gets weaker. Hence for higher temperature, due to weaken the force of attraction between molecule, solid goes into liquid state. and further increase in temperature liquid goes into gaseous state.

To know more about Temperature :

brainly.com/question/11464844

#SPJ2.

Answer:

The tempature changes, and the envronment chnages because of this, therefore making tempature changes in a graph.

Explanation:

sorry if this isnt good

Answer:

A neap tide. Hope this helps

Explanation:

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

            = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m + ½ w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

The extension of the spring in the elevator is 60 mm.

For the extension of the spring to be zero, the elevator must be moving downwards under free fall.

The given parameters;

• mass of the box, m = 5 kg
• extension of the spring, x = 50 mm = 0.05 m

The spring constant is calculated as follows;

F = kx

mg = kx

The tension on the spring in an elevator accelerating upwards is calculated as follows;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 2)

T = 59 N

The extension of the spring is calculated as follows;

For the extension of the spring to be zero, the elevator must be under free fall, such that the tension on the spring is zero.

For free fall, a = g

T = m(g - a) = 0

Learn more here:brainly.com/question/4404276

Answer:

a) the spring will stretch 60.19 mm with the same box attached as it accelerates upwards

b) spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Explanation:

Given that;

Gravitational acceleration g = 9.81 m/s²

Mass m = 5 kg

Extension of the spring X = 50 mm = 0.05 m

Spring constant k = ?

 we know that;

mg = kX  

5 × 9.81 = k(0.05)

k = 981 N/m

a)

Given that; Acceleration of the elevator a = 2 m/s² upwards

Extension of the spring in this situation = X1

Force exerted by the spring = F

we know that;

ma = F - mg

ma = kX1 - mg

we substitute

5 × 2 = 981 × X1 - (5 ×9.81 )

X1 = 0.06019 m

X1 = 60.19 mm

Therefore the spring  will stretch 60.19 mm with the same box attached as it accelerates upwards

B)

Acceleration of the elevator = a

The spring is relaxed i.e, it is not exerting any force on the box.

Only the weight force of the box is exerted on the box.

ma = mg

a = g

a = 9.81 m/s² downwards.

Therefore spring will be relaxed when the elevator accelerates downwards at 9.81 m/s²

Answer:

RMS voltage is 113.1370 V

frequency is 780.685 Hz

voltage is −158.66942 V

maximum current is  2.9739 A

Explanation:

Given data

∆V = 160.0 sin(495t) Volts

so Vmax = 160

and angular frequency = 495

time t = 1/106 s

resistor R = 53.8 Ω

to find out

RMS voltage and frequency of the source and  voltage  and maximum current

solution

we know voltage equation = Vmax sin ωt

here Vmax is 160 as given equation in question

so RMS will be Vmax / √2

RMS voltage = 160/ √2

RMS voltage is 113.1370 V

and frequency = angular frequency / 2π

so frequency = 497 / 2π

frequency is 780.685 Hz

voltage at time (1/106) s

V(t) = 160.0 sin(495/ 108)

voltage = −158.66942 V

so current from ohm law at resistor R 53.8 Ω

maximum current = voltage max / resistor

maximum current =  160 / 53.8

maximum current =  2.9739 A

Final answer:

The root-mean-square voltage of the AC source is 113.14 V, its frequency is 78.75 Hz, and the voltage at time t = 1/106 s is approximately 150.4 V. The current at this peak voltage, when connected to a resistor of 53.8 Ω, is approximately 2.97 A.

Explanation:

The output of an AC voltage source can be represented by the equation V = V₀ sin ωt, where V₀ is the peak voltage, ω is the angular frequency, and t is the time. In this case, V₀ = 160 V and ω = 495 (1/s). The root-mean-square voltage (Vrms), which is commonly used to express AC voltage, can be calculated from the peak voltage using the formula Vrms = V₀/√2 which gives approximately 113.14 V.

The frequency of the source is related to the angular frequency by the equation f = ω/2π, which gives a frequency of approximately 78.75 Hz. To find the voltage at a specific time t = 1/106 s, we substitute these values into the initial equation resulting in V = V₀ sin ωt = approximately 150.4 V.

Finally, the resistance R = 53.8 Ω allows us to calculate the maximum current in the circuit given by I = V/R. The maximum current occurs at the peak voltage, so I(max) = V₀/R = approximately 2.97 A.

Learn more about AC Voltage Source here:

brainly.com/question/32877157

#SPJ3

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt =  λ / 2

t =  λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .