A lion chasing a zebra sprints 200 meters in ten seconds what is the lions average speed

A cylinder 0.170 m in diameter rotates in a lathe at 530 rpm . what is the tangential speed of the surface of the cylinder?

Ad libitum [116K]

Diameter = 0.170 meter
Circumference = 0.170 π meters

530 rpm = 530 circumferences / minute

   = (530 x 0.170 π meters) / minute

   =    283.06 meter.minute

   =    4.72 meters/second

8 0

3 years ago

Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

$m_1=2.2\times 10^{-8} kg$

$m_2=4.8\times 10^{-8} kg$

same charge on both masses

potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

$qvB=\frac{mv^2}{r}$

$B=\frac{\sqrt{\frac{2Vm}{q}}}{r}$

and B is equal for both particles

thus $\frac{m}{r^2}=constant$

$\frac{m_1}{r_1^2}=\frac{m_2}{r_2^2}$

$r_2^2=\frac{4.8}{2.2}\times 8^2$

$r_2=11.81 cm$

4 0

3 years ago

A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th

defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0

3 years ago

A ball rolls off the end of a horizontal table that is 4 meters off the ground. It is measured that the ball lands 3 meters away

ElenaW [278]

Answer:

The speed at which the ball rolled off the end of the table is 3.3 m/s

Explanation:

Hi there!

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the table.

The position vector of the ball can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

r = position vector.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity.

When the ball reaches the ground, its position will be:

r final = (3, -4)

Then:

3 = x0 + v0x · t

-4 = y0 + v0y · t + 1/2 · g · t²

Since the origin of the frame of reference is located at the edge of the table, x0 and y0 = 0. v0y is also 0 ( see the initial velocity vector in the figure to elucidate why). Then:

3 m = v0x · t

-4 m = 1/2 · g · t²

We can solve for "t" in the equation of the y-component and use it in the equation of the x-component to obtain v0x:

-4 m = 1/2 · g · t²

-4 m = -1/2 · 9.8 m/s² · t²

8 m / 9.8 m/s² = t²

t = 0.9 s

Then:

3 m = v0x · 0.9s

3 m/ 0.9 s = v0x

v0x = 3.3 m/s

The speed at which the ball roll off the end of the table is 3.3 m/s

8 0

3 years ago

A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top?

kati45 [8]

Answer:

8361.063 J

Explanation:

7 0

2 years ago