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Svetlanka [38]
2 years ago
12

A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed

Physics
2 answers:
den301095 [7]2 years ago
6 0

Total distance traveled = 250m

Total time taken = 30 seconds

Now,

Average speed = total distance

__________

time taken

= 250 m ÷30 second

= 8.3m/s

Kaylis [27]2 years ago
5 0

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

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If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0
3 years ago
15.One girl was involved in a bus crash. She told the police that she was seated in the last line of seats and that when the bus
Alona [7]

Answer: It is not likely.

Explanation:

When the bus is moving forward, all the objects inside of it also are moving forward.

Now, as the objects inside the buss are not fixed to the bus, if the bus suddenly stops the objects inside of it will keep moving forward, because of the conservation of the momentum, defined as the quantity of motion (Similar to when you are in a car and it suddenly stops, you can feel the forward impulse).

Then is not likely that, in a case where the bus stops suddenly, an object inside the bus flies backward in opposite direction to the previous movement of the bus.

5 0
3 years ago
Nuclei of u-238 atoms are
Elis [28]
 <span>Nuclei of u-238 atoms are </span><span>unstable and spontaneously emit alpha particles.</span>
5 0
3 years ago
The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial
zlopas [31]

Answer:

2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

1/[A] = kt + 1/[A°]

Where,

A = concentration remaining at time, t

A° = initial concentration

k = rate constant

1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81

1/[A] = 0.345

= 1/0.345

= 2.9 M.

6 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
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