Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Answer:
carbon has four unpaired electrons in its valence shell . hydrogen having one unpaired electron in its valence shell comes to bond with carbon by sharing a pair of electrons .since carbon needs 4 electrons to be stable, 4 hydrogen atoms take part in the bond . It's a covalent bond because the difference between the electronegativity of carbon and hydrogen is quite small .
The MA is 6! Hope This Helps!
Answer:
2.64 x 10⁻⁶T
Explanation:
The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;
B = (μ₀ I) / (2π r) ----------------(i)
B is magnetic field
I is current through the wire
r is the distance from the wire
μ₀ is the magnetic constant = 4π x 10⁻⁷Hm⁻¹
From the question;
I = 0.7A
r = 0.053m
Substitute these values into equation (i) as follows;
B = (4π x 10⁻⁷ x 0.7) / (2π x 0.053)
B = 2.64 x 10⁻⁶T
Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T
Answer:
<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>
- <u><em>Initial velocity = 33 m/s</em></u>
- <u><em>Final velocity = 0 m/s</em></u>
- <u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>
Explanation:
- <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
- <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
- <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
