A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>
m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction
Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>
Net force = √625 = 25N
F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2
Answer: 5.00 m/s^2
b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal.
</span>
<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction
Components of F2
F2,x = F2*cos(60) = 15N / 2 = 7.5N
F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N
Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N
Total force in y = F2,y = 13.0 N
Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =
= 30.42N
a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2
Answer: 6.08 m/s^2
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Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
Hence the velocity of the particle at time t is 
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second
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!!!
The variation of momentum (= the impulse) of the car during the impact is
does not change whether the car has an airbag or not, because
and 1) the mass of the car is always the same 2) the change in velocity of the car is always the same,
so if
is constant and F is reduced by a factor 110, then
(the duration of the collision) must be increased by a factor 110 with the airbag.
