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2 years ago

A train travels 190km in 3.0 hours and then 120 km in 2.0 hours. What is it’s average speed ?

Physics

1 answer:

xxMikexx [17]2 years ago

6 0

Answer:

60 km/h

Explanation:

Simplify the speed:

120÷2=60

Hence, the average speed is 60 km/h.

You might be interested in

Using Rayleigh's criterion, what is the smallest separation between two pointlike objects that a person could clearly resolve at

atroni [7]

Answer:

y = 52.44 10⁻⁶ m

Explanation:

It is Rayleigh's principle that two points are resolved if the maximum of the diffraction pattern of one matches the minimum the diffraction pattern of the other

Based on this principle we must find the angle of the first minimum of the diffraction expression

a sin θ= m λ

The first minimum occurs for m = 1

sin θ = λ / a

Now let's use trigonometry the object is a distance L = 0.205 m

tan θ = y / L

Since the angles are very small, let's approximate

tan θ = sin θ/cos θ = sin θ

sin θ = y / L

We substitute in the diffraction equation

y / L = λ / a

y = λ L / a

Let's calculate

y = 550 10⁻⁹ 0.205 / 2.15 10⁻³

y = 52.44 10⁻⁶ m

7 0

3 years ago

Number of complete 90.9 MHz radio waves over a 1.50 km distance

zimovet [89]

You could answer this right away IF you knew the length of each wave, right ?

Well, Wavelength = (speed) / (frequency).

Speed = 3 x 10⁸ m/s (the speed of light)
and
Frequency = 90.9 x 10⁶ Hertz.

So the length of each wave is 3 x 10⁸ / 90.9 x 10⁶ meters.

To answer the question, see how many pieces you have to cut
that 1.5 km into, in order for each piece to be 1 wavelength.
It'll be

(1,500 meters) divided by (3 x 10⁸ meters/sec) / (90.9 x 10⁶ Hz)

To divide by a fraction, flip the fraction and then multiply:

(1500 meters) times (90.9 x 10⁶ Hz)/(3 x 10⁸ meters/sec)

= 454.5

5 0

3 years ago

A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle

s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

$L$ = length of the bat = 81.3 cm = 0.813 m

$m$ = mass of the bat = 0.96 kg

$d$ = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

$T$ = Period of oscillation = 1.35 sec

$I$ = moment of inertia of the bat

Period of oscillation is given as

$T = 2\pi \sqrt{\frac{I}{mgd}}$

$1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}$

$I$ = 0.24 kgm²

6 0

3 years ago

A sky-diver jumps from a stationary balloon. His initial downwards acceleration is 10m/s².

Tatiana [17]

Explanation:

a. The force acting down is gravity, on Earth gravity is 10 m/s^2. When the skydiver jump, the acceleration will start out as -10 m/s^2, but it will eventually equals the air resistance , which is called terminal velocity.

5 0

1 year ago

There are two identical, positively charged conducting spheres fixed in space. The spheres are 40.4 cm apart (center to center)

Nadya [2.5K]

Answer:

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$

Explanation:

$F_1=0.072\ \text{N}$

$F_2=0.115\ \text{N}$

r = Distance between shells = 40.4 cm

$q_1$ and $q_2$ are the charges

$k$ = Coulomb constant = $8.99\times10^{9}\ \text{Nm}^2/\text{C}^2$

Force is given by

$F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow q_1q_2=\dfrac{F_1r^2}{k}\\\Rightarrow q_1q_2=\dfrac{0.072\times 0.404^2}{8.99\times 10^{9}}\\\Rightarrow q_1q_2=1.307\times 10^{-12}\\\Rightarrow q_1=\dfrac{1.307\times 10^{-12}}{q_2}$

$F_2=\dfrac{kq^2}{r^2}\\\Rightarrow q=\sqrt{\dfrac{F_2r^2}{k}}\\\Rightarrow q=\sqrt{\dfrac{0.115\times 0.404^2}{8.99\times 10^{9}}}\\\Rightarrow q=1.44\times 10^{-6}\ \text{C}$

$q=\dfrac{q_1+q_2}{2}\\\Rightarrow q_1+q_2=2q\\\Rightarrow q_1+q_2=2\times1.44\times 10^{-6}\\\Rightarrow q_1+q_2=2.88\times 10^{-6}$

Substituting the above value of $q_1$ we get

$\dfrac{1.307\times 10^{-12}}{q_2}+q_2=2.88\times 10^{-6}\\\Rightarrow q_2^2-2.88\times 10^{-6}q_2+1.307\times 10^{-12}=0\\\Rightarrow \frac{-\left(-0.00000288\right)\pm \sqrt{\left(-0.00000288\right)^2-4\times \:1\times \:1.307\times 10^{-12}}}{2\times \:1}\\\Rightarrow q_2=2.32\times 10^{-6}, 5.64\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{2.32\times 10^{-6}}\\\Rightarrow q_1=5.63\times 10^{-7}$

$q_1=\dfrac{1.307\times 10^{-12}}{q_2}=\dfrac{1.307\times 10^{-12}}{5.64\times 10^{-7}}\\\Rightarrow q_1=2.32\times 10^{-6}$

Since we know $q_1$

$q_1=5.64\times 10^{-7}\ \text{C}$ and $q_2=2.32\times 10^{-6}\ \text{C}$ .

5 0

2 years ago