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Drupady [299]
3 years ago
12

Select the correct answer.

Physics
2 answers:
emmainna [20.7K]3 years ago
7 0

Answer:

A

Explanation:

aliya0001 [1]3 years ago
4 0

Answer:

A. Prophase

Hope this helped :)

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The speed of a cruise ship is 50,000 m/hr. How fast is it moving in m/s?
solong [7]

Answer:

13.888888 (on forever)

Explanation:

there are 3600 seconds in an hour if it is moving at an hourly rate then you just need to divide 50,000 buy 3600 in which case you get 13.88888

3 0
3 years ago
A pickup truck is carrying a 10.0 kgkg toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts
iren2701 [21]

Answer:

Shortest time = 3.47 seconds

Explanation:

We know that Force of gravity of an object is given by;

F_g = mg

Also, Force due to kinetic friction is given as;

F_friction = μ_k•F_n

Force due to static friction is given as;

F_static = μ_s•F_n

Where ;

μ_k = coefficient of kinetic friction

μ_s = coefficient of static friction

F_n = Normal Force

Let's solve for the static crate.

From the free body diagram i attached, the resultant force in the vertical y direction will be;

N - mg = 0

Thus, N = mg

WherN is the

From earlier, we know that;

F_static = μ_s•F_n

Thus,

F_static = μ_s•mg

Now, in the horizontal x direction,

F_static = ma

Thus,

μ_s•mg = ma

m will cancel out, and;

a = μ_s•g

a = 0.3 x 9.81 = 2.943 m/s²

Not, to get the final the final speed of the tool box, we'll use;

V_f = V_i + a•Δt

Where;

Δt is shortest time.

V_f = final velocity

V_i = initial velocity

Thus, making Δt the subject, we have;

Δt = (V_f - V_i)/a = (10.2 - 0)/2.943 = 3.47 s

6 0
2 years ago
A golf ball is hit so that it leaves the club face at a velocity of 45m/s at an angle of 40° to the horizontal. by ignoring the
tatuchka [14]

Answer:a)45cos40 b)45sin40 c) 2.95 d)102

Explanation:

a) v_{x} = 45 cos40

b) v_{y}=45sin40

c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -<em>g</em>. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: v_{f} = v_{i} + at.

0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

x=vt\\x=45cos40(2.95) =102m

6 0
2 years ago
Particle q1 has a positive 6 µc charge. particle q2 has a positive 2 µc charge. they are located 0.1 meters apart. recall that k
Dmitrij [34]

(a) Force between the two charges

The electrostatic force between the two charges is given by:

F=k\frac{q_1 q_2}{r^2}

where k is the Coulomb's constant, q1 and q2 the two charges, r their separation.


In this problem:

q_1 =6 \mu C=6 \cdot 10^{-6}C

q_2=2 \mu C=2 \cdot 10^{-6}C

r=0.1 m


Substituting into the equation, we find

F=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(6 \cdot 10^{-6}C)(2 \cdot 10^{-6}C)}{(0.1 m)^2}=10.8 N


(b) direction of particle q2

Particle q2 wants to move in the direction of the force acting on it. The direction of the force depends on the relative sign of the two charges: like charges attract each other, opposite charges repel each other. In this case, the two charges are both positive, so they repel each other and q2 tends to move away from particle q1.

7 0
3 years ago
Read 2 more answers
Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral
cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ      

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂)   (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

6 0
3 years ago
Read 2 more answers
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