Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration, 
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.
Answer:
Specific gravity of other fluid = .854 (Approx)
Explanation:
Given:
Mass of water = 35 g
Mass of filled bottle with water = 98.44 g
Mass of filled bottle with fluid = 89.22 g
Computation:
Mass of water = 98.44g - 35g = 63.44g
Density of water = 1000 g/L
Volume of bottle = 63.44/1000 = 0.06344L
Mass of other liquid = 89.22g - 35g = 54.22g
Density of other liquid = 54.22g/0.06344L = 854.665826 g/L
Water has a specific gravity = 1
So , specific gravity of other fluid
1000 / 854.665826 = 1 / specific gravity of other fluid
Specific gravity of other fluid = .854 (Approx)
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
Answer:
<em>d. unchanged.</em>
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Answer:
To calculate anything - speed, acceleration, all that - we need <em>data</em>. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to <em>measure </em>it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring <em>time </em>and <em>distance</em>, and a method for measuring each.
Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:
- Press start on the stopwatch right as you pass a lamp post
- Each time you pass another lamp post, press the lap button on the stopwatch
- Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
- Record your data
- Calculate the time intervals for passing each lamp post using the lap data
- Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed
Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.
