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3 years ago

A bus travels 30km in 1/2h. What is it's average speed in km/h

1 answer:

5 0

If a bus travels 30 km in 1/2 hr, then in one hr, he can travel twice the distance.
30*2=60 km

Final answer: 60 km per hr

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(Equation 17.6) Write the equation for the path-length difference at a bright fringe (constructive interference). Define all var

Andrews [41]

Answer:

d sin tea = m λ

Explanation:

When we have a two-slit system, the optical path difference determines whether the intensity reaching an observation screen is maximum or zero.

To find this difference in optical path, we assume that the screen is much farther than the gap is, we draw a perpendicular from ray 1 to the second ray

OP = d sin θ

now to have constructive interference and see a bright line this leg must be an integer number of wavelengths, ose

d sin tea = m λ

where

d is the distance between the two slits

θ complexion the angle sea the point hold it between the two slits

λ the wavelength of the coherent light used

m an integer, which counts the number of lines of interference

Units in the SI system

d, lam in meters

θ degrees

m an integer

7 0

2 years ago

A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

NISA [10]

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity $g=9.8m/sec^2$

Coefficient of static friction $\mu _s=0.584$

Coefficient of kinetic friction $\mu _k=0.399$

(a) Static friction force is given by $F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N$

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by $F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N$

3 0

2 years ago

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

8 0

3 years ago

If it requires 3.0 J of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re

-BARSIC- [3]

Answer:

Work done = 13605.44

Explanation:

Data provided in the question:

For elongation of 2.1 cm (0.021 m) work done by the spring is 3.0 J

The relation between Energy (U) and the elongation (s) is given as:

U = $\frac{1}{2}kx^2$ ................(1)

where,

k is the spring constant

on substituting the valeus in the above equation, we get

3.0 = $\frac{1}{2}k\times0.021^2$

k = 13605.44 N/m

now

for the elongation x = 2.1 + 4.1 = 6.2 cm = 0.062 m

using the equation 1, we have

U = $\frac{1}{2}\times13605.44\times (0.062)^2$

U = 26.149 J

Also,

Work done = change in energy

W = 26.149 - 3.0 = 23.149 J

4 0

2 years ago

Just need help with 1 and 2 please :D i’m having a bit of trouble :/

dexar [7]

1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.

7 0

3 years ago