If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?

An object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in

CaHeK987 [17]

Answer:

a.) 10Hz

b.) 0.1 s

c.) 187.4 m/s

d.) -412.6 m/s^2

Explanation:

Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.

(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?

from the equation given, the angular speed w = 20π

but w = 2πf

where f = frequency.

substitute w for 20π

20π = 2πf

f = 20π/2π

f = 10 Hz

(b) What is t the first time that the object is at its farthest right?

since F = 1/T

T = 1 / f

T = 1/10

T = 0.1 s

Therefore, the t of first time that the object is at its farthest right is 0.1 s

(c) At the time found in part (b), what is the object's velocity?

The velocity can be found by differentiating the equation;

x(t) = 3sin(20πt)

dx/dt = 60πcos(20πt)

where dx/dt = velocity V

V = 60πcos(20π * 0.1)

V = 187.4 m/s

(d) At the time found in part (b), what is the object's acceleration?

to get the acceleration, differentiate equation V = 60πcos(20πt)

dv/dt = -1200πSin(20πt)

dv/dt = acceleration a

a = -1200πSin(20πt)

substitute t into the equation

a = -1200πSin(20π * 0.1)

a = - 412.6 m/s^2

8 0

2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

g100num [7]

Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

6 0

2 years ago

If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?

barxatty [35]

Answer:

$2.5\cdot 10^{-4}$

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

$S=\frac{\Delta L}{L_0}$

where

$\Delta L$ is the change in length of the object

$L_0$ is the original length of the object

In this problem, we have $L_0 = 4 m$ and $\Delta L=1 mm=0.001 m$ , therefore the strain is

$S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}$

5 0

2 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

2 years ago

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0

1 year ago