The diameter of circular park is 80 m find its area

Question 1 (1 point)

MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work $W$ done by the frictional force is

$W= Fd$ .

Now, $F = 60N$ and $d =10m$ ; therefore,

$W= (60N)(10m)$

$\boxed{W = 600J.}$

Hence, the work done by friction is 660J.

7 0

3 years ago

A car traveling 34 mi/h accelerates uniformly for 4 s, covering 615 ft in this time. What was its acceleration? Round your answe

Contact [7]

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

$34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2$

Acceleration is 51.94 ft/s²

$v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s$

Final velocity at this time is 257.63 ft/s

5 0

3 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

3 years ago

In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,

taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

$f = \frac{1}{2\pi \sqrt{LC} }$

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

$f' = \frac{1}{2\pi \sqrt{2L2C} }$

$f' = \frac{1}{2\pi \sqrt{4LC} }$

$f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}$

$f' = \frac{1}{2} f$

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

brainly.com/question/13040523

#SPJ4

6 0

2 years ago

Convert 1nanosecond in to its SI init

SOVA2 [1]

Convert 1nanosecond in to its SI init

In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.

In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)

8 0

3 years ago

The diameter of circular park is 80 m find its area​

The diameter of circular park is 80 m find its area