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3 years ago

Time management skills include

Physics

1 answer:

dexar [7]3 years ago

7 0

D. Budgeting time, avoiding stress, and prioritizing.

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Which student is doing work?

Vilka [71]

Answer:

hope this helps

Explanation:

8 0

3 years ago

Read 2 more answers

A student conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. In trial 1,

Harlamova29_29 [7]

Answer:

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

F = m a

a = F / m

We use kinematics to find lips

v = v₀ + a t

v = v₀ + (F / m) t

The moment is defined by

p = m v

The moment change

Δp = m v - m v₀

Let's replace the speeds in this equation

Δp = m (v₀ + F / m t) - m v₀

Δp = m v₀ + F t - m v₀

Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

8 0

3 years ago

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

3 years ago

While skateboarding at 19 km/h, Alana throws a tennis ball at 11 km/h to her friend Oliver. If Alana is the reference frame, the

kifflom [539]

Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.

4 0

3 years ago

You read in a science magazine that on the Moon, the speed of a shell leaving the barrel of a modern tank is enough to put the s

olga_2 [115]

To solve this problem we will use the definition of the kinematic equations of centrifugal motion, using the constants of the gravitational acceleration of the moon and the radius of this star.

Centrifugal acceleration is determined by

$a_c = \frac{v^2}{r}$