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2 years ago

Explain how to find the acceleration of an object that has one-dimensional horizontal motion.

1 answer:

7 0

Acceleration is how much the velocity changes within a period of time so,

Acceleration= is the change in velocity divided by change in time

your units will be m/s squared

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A 1.0 kg rock moving at 8.4 m/s will have ______ of kinetic energy

BlackZzzverrR [31]

K.E=1/2mv^2 K.E=1/2multiply1multiply8^2=32joules

4 0

3 years ago

THE ELECTROMAGNET SPECTRUM ONLY HAS WHICH SHAPE OF WAVE?

brilliants [131]

Answer: Transverse

Explanation: Transverse waves possess a vertical wave motion and a horizontal particle motion.

4 0

3 years ago

A space station in the shape of a 100 m-diameter (50m radius) wheel is spinning so as to impart a linear tangential speed of 22.

zalisa [80]

Answer:

correct option is b. 31.3 m/s

Explanation:

given data

artificial gravity a1 = 1 g

artificial gravity a2 = 2 g

diameter = 100 m

radius r= 50 m

speed v1 = 22.1 m/s

solution

As acceleration is ∝ v²

so we can say

$\frac{a2}{a1} = \frac{v2}{v1}$ .....................1

put here value

$\frac{2}{1} = \frac{v2}{22.1}$

solve it

v2 = $\sqrt{2 }$ × 22.1

v2 = 31.25 m/s

so correct option is b. 31.3 m/s

4 0

3 years ago

Estimate the moment of inertia of a bicycle wheel 66 cm in diameter. The rim and tire have a combined mass of 1.2 Kg

Fed [463]

<span>For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr^2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses. So the I = (1.2 kg)(0.66m/2)^2 = 0.1307 kg m2</span>

3 0

3 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

The entropy change of the rod will be zero.

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

3 years ago