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sertanlavr [38]
3 years ago
15

What is the best inference for the height of the plant after 35 hours?

Physics
1 answer:
natima [27]3 years ago
5 0

Option D, 1.95 cm

1.8+2.1=3.9/2 = 1.95

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An object at rest does not _____ and an object in motion does not _____, unless an _____ force acts upon it
mel-nik [20]

Answer:

An object at rest does not move and an object in motion does not change its velocity, unless an external force acts upon it

Explanation:

This statement is also known as Newton's first law, or law of  inertia.

It states that the state of motion of an object can be changed only if there is an external force (different from zero) acting on it: therefore

- If an object is at rest, it will remain at rest if there is no force acting on it

- If an object is moving, it will continue moving at constant velocity if there is no force acting on it

This phenomenon can be also understood by looking at Newton's second law:

F = ma

where

F is the net force on an object

m is the mass

a is the acceleration

If the net force is zero, F = 0, the acceleration of the object is also zero, a = 0: therefore, the velocity of the object does not change, and it will continue moving at the same velocity (which can be zero, if the object was at rest).

5 0
3 years ago
A mass of 2.1 kg the ball leaves her hand with the speed of 30 mass find the energy of the ball
snow_lady [41]
Ke= 1/2 x m x v^2
Ke= 1/2 x 2.1 x 30^2
Energy = 945 J
8 0
3 years ago
outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp
sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s

So, the speed of the box is 0.489 m/s.

3 0
3 years ago
Calculate the force experienced by a positive test charge of 8.1 × 10-9 coulombs if the electric field strength is 9.4 × 107 new
Arte-miy333 [17]
The electric force exerted by an electric field of intensity E on a charge q is equal to the product between E and q, so:
F=qE=(8.1 \cdot 10^{-9}C)(9.4 \cdot 10^7 N/C)=0.76 N
5 0
3 years ago
Read 2 more answers
Two charged particles are placed 2.0 meters apart. The first charge is +2.0 E-6 C, and the second charge is +4.0 E-6 C. What is
ipn [44]
F = kq1q2/r^2 

<span>q1 is first charge </span>
<span>q2 is second charge </span>
<span>k is 9 E9 </span>
<span>r is distance between them </span>

<span>F = (9E9)(2 E-6)(4 E-6)/2^2 = 0.018 N </span>

<span>A postive answer indicates a repulsive force</span>
4 0
3 years ago
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