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enyata [817]
3 years ago
6

What is the definition of uniform motion?

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Uniform motion means no part of the motion is changing.

The two parts of motion are . . . speed and direction.

If neither part is changing, that means constant speed in a straight line.


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Can someone please help, ty!
lyudmila [28]

Answer:hypothesis: the plant died of lack  of light,moisture,or water

Explanation:

to test my hypothesis i put a plant in a room at room temp and repeat how i grew it,and observe what went wrong. hope this helps=)

5 0
3 years ago
An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm. The coil lies in the
Darina [25.2K]

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

5 0
3 years ago
A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.
ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

6 0
3 years ago
al aplicar una fuerza de 2 N sobre un muelle este se alarga 4cm.¿cuanto se alargara si la fuerza es el triple?¿que fuerza tendri
3241004551 [841]

1) 12 cm

2) 3 N

Explanation:

1)

The relationship between force and elongation in a spring is given by Hooke's law:

F=kx

where

F is the force applied

k is the spring constant

x is the elongation

For the spring in this problem, at the beginning we have:

F=2 N

x=4 cm

So the spring constant is

k=\frac{F}{x}=\frac{2N}{4 cm}=0.5 N/cm

Later, the force is tripled, so the new force is

F'=3F=3(2)=6 N

Therefore, the new elongation is

x'=\frac{F'}{k}=\frac{6}{0.5}=12 cm

2)

In this second problem, we know that the elongation of the spring now is

x=6 cm

From part a), we know that the spring constant is

k=0.5 N/cm

Therefore, we can use the following equation to find the force:

F=kx

And substituting k and x, we find:

F=(0.5)(6)=3 N

So, the force to produce an elongation of 6 cm must be 3 N.

6 0
3 years ago
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if
sergij07 [2.7K]

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

7 0
2 years ago
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