Answer:
I think c is the answer but I have a little concern on d too
Answer:
The answer is the option a.
Explanation:
We know that magnetic force (Fm) is defined as
Fm = q (v x B)
Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.
"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is
v x B = vBsin (thetha)
So,
Fm = qvBsin (thetha)
And, in case in which v and B are parallel vectors, thetha is zero, and,
sin (thetha)=sin (0) = 0
So, Fm=0
Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
Answer:
P = (2 + 3) * V where V is their initial speed (total momentum)
P = 2 * 10 + 3 * Vx where Vx here would be V3
If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.
Also work depends on the sum of the velocities
W (initial) = 1/2 (2 + 3) V^2 the initial kinetic energy
W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2
It appears that more information is required for this problem
