All categories

2 years ago

What is the car's average velocity (in m/s) in the interval between t = 1.0 s to t = 1.5 s?

Physics

1 answer:

natali 33 [55]2 years ago

4 0

Answer:

1.4 m/s

Explanation:

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

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Velocity (v) =..?

The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:

Velocity = change of displacement /time

v = Δd / Δt

Thus, with the above formula, we can obtain the velocity of the car as follow:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9

= 0.7 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Change in time (Δt) = t2 – t1

= 2 – 1.5

= 0.5 s

Velocity (v) =..?

v = Δd / Δt

v = 0.7/0.5

v = 1.4 m/s

Therefore, the velocity of the car is 1.4 m/s

You might be interested in

In a single-slit diffraction experiment, a coherent light source illuminates a slit in a barrier, and the resulting pattern is p

UkoKoshka [18]

Answer:

hsjdsdddwqdqwdd

Explanation:

4 0

1 year ago

The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie

Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I' and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0

2 years ago

Read 2 more answers

pls help quick. the number line shows the starting and ending velocities for ball 1 what's the change in velocity of ball 1 calc

pentagon [3]

Answer:

The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.

The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds

7 0

2 years ago

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the

Taya2010 [7]

Answer:

The angular velocity is 15.37 rad/s

Solution:

As per the question:

$\theta = 54.6^{\circ}$

Horizontal distance, x = 30.1 m

Distance of the ball from the rotation axis is its radius, R = 1.15 m

Now,

To calculate the angular velocity:

Linear velocity, v = $\sqrt{\frac{gx}{sin2\theta}}$

v = $\sqrt{\frac{9.8\times 30.1}{sin2\times 54.6}}$

v = $\sqrt{\frac{294.98}{sin109.2^{\circ}}} = 17.67\ m/s$

Now,

The angular velocity can be calculated as:

$v = \omega R$

Thus

$\omega = \frac{v}{R} = \frac{17.67}{1.15} = 15.37\ rad/s$

8 0

2 years ago

A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the

k0ka [10]

<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>

8 0

2 years ago