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Explanation:
We'll use the free fall formula, which is,
y = 0.5*g*t^2
where
t = time in secondsg = acceleration of gravity (in m/s^2)y = vertical distance the object has fallen (in meters)In this case,
t = 4 secondsg = 10 m/s^2 approximatelyNote: The freefall assumes the object is not thrown down, and simply dropped (so initial velocity is 0).
So,
y = 0.5*g*t^2
y = 0.5*10*4^2
y = 0.5*10*16
y = 5*16
y = 80
The object has fallen about 80 meters. This value is approximate because g = 10 is approximate. Use g = 9.81 for a more accurate value (though even then this value of g isn't exact either). In this case, your teacher has opted to use g = 10.
Another reason why 80 meters is an approximate value is because we haven't accounted for air resistance. As the object falls through the air, the air particles push up and resist the movement of the falling object, and therefore slow it down. Think of a parachute. For entry level physics courses, you'll ignore air resistance as it greatly complicates the problem. Though it's handy to keep it in mind for later physics courses down the road.
A circuit consists of a 12 V battery connected across a single resistor. If the current in the circuit is
3 A, calculate the size of the resistor
Answer:
4 Ohms
Explanation:
Apply the formula:
Voltage = I (current) . Resistance
You can change it the way you want to use for your purpose.
In this case...
R = V/I
R = 12/3
R = 4 Ohms (Ohm is the unit of measurement of eletrical resistance)
Calculate the work done in lifting 200 kg of water through a vertical height of 6 metres
(g = 10 m/s)
(A) 5000 J
(B) 12000 J
(C) 25000 J
(D) 15000 J
Answer:
[tex]\bf\pink{(C)\:12000\:J}[/tex]Explanation:
Given :-[tex]\sf\red{Mass = 200 \ kg}[/tex][tex]\sf\orange{Gravity = 10 \ m/s}[/tex][tex]\sf\green{Height = 6 \ m}[/tex]Need to find :-[tex]\sf\blue{Work \ done}[/tex]Formula required :-[tex]\sf\purple{Work \ done = Mass \times Gravity \times Height}[/tex]Solution :-[tex]\to\:\:\sf\red{Work \ done = Mass \times Gravity \times Height}[/tex]
[tex]\to\:\:\sf\orange{Work \ done = 200 \times 10 \times 6}[/tex]
[tex]\to\:\:\sf\green{Work \ done = 2000 \times 6}[/tex]
[tex]\to \:\ \sf\blue{ Work \ done = {\bf{\blue{1200\:J}}}}[/tex]
Hence, [tex]\bf\green{(B)}[/tex] is the correct option.You (50 kg) are standing on a skateboard when your friend throws a 5 kg ball at 10 m/s towards your head. You catch the ball to save your face. Describe your motion after you catch the ball, use numbers.
Answer:
v₁ = 1 [m/s].
Explanation:
This problem can be solved by using the principle of conservation of linear momentum. Where momentum is conserved before and after catching the snowball.
[tex]P=m*v[/tex]
where:
P = linear momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
[tex]P=m*v\\P=5*10\\P=50[m/s][/tex]
As the momentum is conserved, it will be the same as that received by the person who catches the snowball.
[tex]50= 50*v_{1}\\v_{1}=1[m/s][/tex]
That is, the person who catches the snowball moves backwards with a speed of 1 [m/s].