You can find
1) time to hit the ground
2) initial velocity
3) speed when it hits the ground
Equations
Vx = Vxo
x = Vx * t
Vy = Vyo + gt
Vyo = 0
Vy = gt
y = yo - Vyo - gt^2 / 2
=> yo - y = gt^2 / 2
1) time to hit the ground
=> 8.0 = g t^2 / 2 => t^2 = 8.0m * 2 / 9.81 m/s^2 = 1.631 s^2
=> t = √1.631 s^2 = 1.28 s
2) initial velocity
Vxo = x / t = 6.5m / 1.28s = 5.08 m/s
3) speed when it hits the ground
Vy = g*t = 9.81 m/s * 1.28s = 12.56 m/s
V^2 = Vy^2 + Vx^2 = (12.56 m/s)^2 + (5.08 m/s)^2 = 183.56 m^2 / s^2
=> V = √ (183.56 m^2 / s^2) = 13.55 m/s
1.25, if the field is 1 meter away. I think that that is correct.
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Answer:
Explanation:
Point P is situated outside sphere . For a non conducting sphere , electric field outside sphere is given by the relation
E = k Q / d²
15000 = 9 x 10⁹ x Q / .5²
Q = 416.67 X 10⁻⁹
Maximum field will be when d = r ( radius )
maximum E = kQ / r²
= 9 x 10⁹ x 416.67 X 10⁻⁹ / .3²
= 41667 N / C
Answer:
there are 3.018 kcal= 3018 cal per gram of candy
Explanation:
If the assume that the calorimeter is perfectly insulated, then all the heat released by the combustion is absorbed by the calorimeter.
Also knowing that Q= C * ΔT , where C= heat capacity of the calorimeter , ΔT= temperature change , Q = heat released by the combustion of the candy
replacing values
Q = C * ΔT = 33.90 kJ/°C * 2.46°C = 83.394 kJ
since Q is the heat released when burned all the mass m of the candy, the number of calories per gram of candy will be
q = Q/m =83.394 kJ / 6.60 g = 12.635 kJ/g
q = 12.635 kJ/g * 1 kcal / 4.186 kJ = 3.018 kcal per gram of candy
