Ask question

Ask question

All categories

3 years ago

11

A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

nd the 80.00 m radius circular track. What is the frequency of the race car's circular motion?

1 answer:

koban [17]3 years ago

3 0

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

You might be interested in

Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a curre

valina [46]

Answer:

3 x 10^-4 N/m

Attractive

Explanation:

r = 0.1 m, i1 = 10 A, i2 = 15 A

The force per unit length between two parallel wires is given by

$F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}$

$F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}$

F = 3 x 10^-4 N/m

Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.

3 0

3 years ago

Explain our ideas about why the terrestrial planets are rocky and have less gas than the giant planets.

snow_lady [41]

Answer: The terrestrial planets are mainly comprised of silicate minerals and also there is presence of some metals, which were subsequently formed in the geological time scale, after the accumulation of smaller dust particles, due to the gravitational pull, during the time of formation of the planets.

Due to the location of these terrestrial planets near to the sun, their surface temperature is high enough to hold the gases like hydrogen and helium. Thus, they have a high density but less gases compared to the outer planets.

3 0

3 years ago

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Gekata [30.6K]

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = $6.5*10^{4}\ m/s$

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity $[\theta]$ = 15 degree.

We are asked to calculate the magnetic force experienced by the charged particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force $\vec F=q(\ \vec V \times \vec B)$

i.e F = $=qVBsin\theta$

$=6.8*6.5*10^{4}*1.4*sin15$

$=61.88*10^{4}sin15$

$=61.88*10^4*0.2588\ N$

$=16.016*10^4\ N$

$=1.6*10^5\ N$

Hence, the force experienced by the charged particle is C i.e $1.6*10^5 N$

4 0

3 years ago

Read 2 more answers

In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh

nata0808 [166]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

$Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}$

Where,

$C_d$ is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = $\frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2$

A₂ = Area at the throat

A₂ = $\frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2$

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

$Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}$

or

$Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}$

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

7 0

3 years ago

Describe the sequence of events in the lithification of a sandstone

guajiro [1.7K]

Lithification is a diagenetic process in which loose sediment is converted to hard rocky compaction and cementation. So sediments (sand) are buried the increase in pressure from the weight of the overlaying material pushes the grains closer together. The volume of sediment is reduced and the fluids between the grains are also squeezed out. This leaves the sandstone tightly compressed <span>together this is lithification.</span>

6 0

3 years ago

Read 2 more answers