Natural history research involves ____ unobsevable and unrepeatable past events.

A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate

maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

$\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}$

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

$a=\frac{11.7-26.5}{5.2}=-2.85m/s^2$

Angular acceleration

$\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2$

Angular acceleration = -3.95 rad/s²

Magnitude = 3.95 rad/s²

4 0

3 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

A student releases a marble from the top of a ramp. The marble increases

Fed [463]

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at

Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)

7 0

3 years ago

You are investigating an elevator accident which happened in a tall building. An elevator in this building is attached to a stro

Advocard [28]

Answer:

a) F = 2250 Ib

b) F = 550 Ib

c) new max force ( F newmax ) = 2850 Ib

Explanation:

A) The force the wall of the elevator shaft exert on the motor if the elevator starts from rest and goes up

max capacity of elevator = 24000 Ibs

counterweight = 1000 Ibs

To calculate the force (F) :

we first calculate the Tension using this relationship

Counterweight (1000) - T = ( 1000 / g ) ( g/4 )

Hence T = 750 Ib

next determine F

750 + F - 2400 = 2400 / 4

hence F = 2250 Ib

B ) calculate Tension first

T - 1000 = ( 1000/g ) ( g/4)

T = 1250 Ib

F = 2400 -1250 - 2400/ 4

F = 550 Ib

C ) determine design limit

Max = 2400 * 1.2 = 2880 Ib

750 + new force - 2880 = 2880 / 4

new max force ( F newmax ) = 2850 Ib

8 0

3 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?

12345 [234]

   Density = (mass) / (volume)

      4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by 0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

   mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

   = (360 kg) x (9.8 m/s²)

   = (360 x 9.8) kg-m/s²

   = 3,528 newtons .

That's the force of gravity on this block, and it doesn't matter
what else is around it. It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity. That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water. Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water. The weight
of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

   (3,528 - 882) = 2,646 newtons (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water". The force of gravity doesn't change, but
there's another force, working against gravity, in the water.

5 0

3 years ago

Read 2 more answers