It is not possible to accurately calculate instantaneous velocity.<br> TRUE<br> FALSE

A pendulum has a period of 1.69 s on earth. what is its period on mars, where the acceleration of gravity is about 0.37 that on

love history [14]

The formula for the period of the pendulum is
$T=2 \pi \sqrt{ \frac{L}{g} }=$
where L is the pendulum's length and g the gravitational acceleration.

Labeling with E the Earth and with M Mars, we can write the period of the pendulum on Earth as
$T_E = 2 \pi \sqrt{ \frac{L}{g_E} }$
and the period of the pendulum on Mars as
$T_M = 2 \pi \sqrt{ \frac{L}{g_M} }$

if we calculate the ratio, we get
$\frac{T_M}{T_E}= \sqrt{ \frac{g_E}{g_M} }$
but we know that the gravitational acceleration on Mars is 0.37 times the gravitational acceleration on Earth:
$g_M = 0.37 g_E$
Substituting into the formula, we find
$\frac{T_M}{T_E}= \sqrt{ \frac{1}{0.37} }=1.64$
And so, the period of the pendulum on Mars is
$T_M = 1.64 T_E = 1.64 (1.67 s)=2.78 s$

8 0

3 years ago

Read 2 more answers

Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the

nikklg [1K]

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673 / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr = $\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }$

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

so

pressure = Pr × Pc

pressure = 0.57 × 22064 × 10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

so

pressure P is

$\frac{0.02 - 0.020030}{0.015671 - 0.020030}$ = $\frac{ P - 12.5}{15 - 12.5}$

so

Pressure = 12517 kPa

4 0

3 years ago

A stone is launched vertically upward from a cliff 384 ft above the ground at a speed of 80 ft divided by s. Its height above th

goldenfox [79]

Answer:

2.5 seconds

Explanation:

s(t) = -16t^2 + 80t + 384

for

0≤t≤8

First we differentiate s(t) to get s'(t)

s'(t) = -32t + 80

Let us then find the critical point; thus we will equate s'(t) to zero and then search for values where s'(t) is undefined

s'(t) = -32t + 80 = 0

t = 80/32

t = 2.5 sec

Let us evaluate s at the critical points and end points

s(0) = -16(0)^2 + 80(0) + 384 = 384

s(2.5) = -16(2.5)^2 + 80(2.5) + 384 = 684

s(8) = -16(8)^2 + 80(8) + 384 = 0

Thus, the stone attains it maximum height of 684ft at at t=2.5s

3 0

4 years ago

If a book has a a mass of 2 kg, how much does the book weigh?

Darina [25.2K]

Answer:

A. 20 N

Explanation:

weight of book is 2×10=20N .....

5 0

3 years ago

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for

nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ( $W$ ), measured in newtons, is:

$W = m\cdot g$ (1)

Where:

$m$ - Mass of the block of ice, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $m = 14\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , the magnitudes of the weight and normal force of the block of ice are, respectively:

$N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$N = W = 137.298\,N$

And the pull force is:

$F_{pull} = 98\,N$

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

$m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}$ (2)