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Klio2033 [76]
3 years ago
15

A displacement vector is 34.0 m in length and is directed 60.0° east of north. What are the components of this vector? Northward

Eastward choice component component 1 29.4 m 17.0 m 2 18.2 m 28.1 m 3 22.4 m 11.5 m 4 17.0 m 29.4 m 5 25.2 m 18.2 m A) Choice 1 B) Choice 2 C) Choice 3 D) Choice 4 E) Choice 5 79) If I walk 8.0 meters in a straight line and then walk 5.0 meters in another straight line, the total displacement cannot have a magnitude of A) 8 m. B) 2 m. C) 13 m. D) 5 m. E) 3 m.
Physics
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

A)  29.4 m 17.0 m; B) 2 m

Explanation:

If a vector is 34.0 m in length and is directed 60.0° east of north (which means 30.0° over the horizontal), then its coordinates will be:

Horizontal: (34.0 m)cos(30.0°)=29.4 m

Vertical: (34.0 m)sin(30.0°)=17 m

If one person walks 8.0 meters in a straight line and then walks 5.0 meters in another straight line, then the minimum displacement would be to go back over his tracks, displacing himself 8m-5m=3m, while the maximum displacement would be going straight ahead, displacing himself 8m+5m=13m. Any answer outside this interval is impossible (2m).

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Newtons law of motion for every action there’s an equal and opposite reaction.
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A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
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Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

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3 years ago
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An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
3 years ago
A skydiver of mass 80kg jumps from a slow moving aircraft and reach a terminal speed of 50 m/s. What's her acceleration when her
miskamm [114]

Answer:

a = g = 9.81[m/s^2]

Explanation:

This problem can be solve using the second law of Newton.

We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.

m*g = m*a

where:

g = gravity = 9.81[m/s^2]

a = acceleration [m/s^2]

Note: If the skydiver will be under air resistance forces his acceleration will be different.

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Suppose a runner completes one lap around a 400-m track in a time of 50 s. Calculate the average speed of the runner.
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Answer:

8m/s

Explanation:

Average Speed = distance / time = 400/50 = 8m/s

6 0
3 years ago
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