Answer:
It takes you 32.27 seconds to travel 121 m using the speed ramp
Explanation:
<em>Lets explain how to solve the problem</em>
- The speed ramp has a length of 121 m and is moving at a speed of
2.2 m/s relative to the ground
- That means the speed of the ramp is 2.2 m/s
- You can cover the same distance in 78 seconds when walking on
the ground
<em>Lets find your speed on the ground</em>
Speed = Distance ÷ Time
The distance is 121 meters
The time is 78 seconds
Your speed on the ground = 121 ÷ 78 = 1.55 m/s
If you walk at the same rate with respect to the speed ramp that
you walk on the ground
That means you walk with speed 1.55 m/s and the ramp moves by
speed 2.2 m/s
So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s
Now we want to find the time you will take to travel 121 meters using
the speed ramp
Time = Distance ÷ speed
Distance = 121 meters
Speed 3.75 m/s
Time = 121 ÷ 3.75 = 32.27 seconds
It takes you 32.27 seconds to travel 121 m using the speed ramp
The goal of Science is to expand knowledge.
As we know that range of the projectile motion is given by
here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be
so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates.
Answer:
Option A is correct.
when it is used in a circuit. its terminal voltage will be less than 1.5 V.
Explanation:
The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.
All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.
Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.
But once the interenal resistance (r) of the battery comes into play,
V = I₁ (r + R)
The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)
The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.
Hope this Helps!!!
