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3 years ago

All scientists try to base their conclusions on

Physics

2 answers:

Crank3 years ago

5 0

They base their conclusions off Data that they recorded when conducting experiments.

Dima020 [189]3 years ago

3 0

Mathematics and science

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What is the acceleration of a racing car if it's speed is increased uniformly from 44 m /s to 66 m / s over an 11 second period

jolli1 [7]

Answer:

the acceleration of the race car is 2 m/s²

Explanation:

Given;

initial velocity of the race car, u = 44 m/s

final velocity of the race car, v = 66 m/s

time of motion of the race car, t = 11 s

The acceleration of the race car is calculated as;

$a = \frac{v-u}{t} \\\\a = \frac{66-44}{11} \\\\a = 2 \ m/s^2$

Therefore, the acceleration of the race car is 2 m/s²

7 0

2 years ago

a wire is carrying a 2.45 A current. at what distance from the wire is the magnetic field 1.00x10^-6t

balu736 [363]

Answer:

0.49m

Explanation:

So you need to change the original equation for finding fields to find distance, and then just plug in the numbers

Which equals 0.49meters

Also it was right on Acellus :)

Hope this helps

5 0

3 years ago

An air mass is a large pocket of air with _____.

maw [93]

An air mass is a large pocket of air with a uniform temperature and humidity

5 0

3 years ago

For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km

muminat

Answer with Explanation:

We are given that

$y=(18-2x^2) km$

$x=(4t-3)km$

Differentiate x and y w.r.t t

$\frac{dx}{dt}=4$

$\frac{dy}{dt}=-4x\frac{dx}{dt}$

$\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)$

$v_x=\frac{dx}{dt}=4$

$v_y=\frac{dy}{dt}=-16(4t-3)$

Substitute t=1

$v_x=4$

$v_y=-16(4-3)=-16$

Magnitude of velocity= $\mid v\mid=\sqrt{v^2_x+v^2_y}$

$\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s$

Hence, the magnitude of the missile's velocity=16.49 m/s

$a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0$

$a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64$

Substitute t=1

$a_x=0,a_y=-64$

$\mid a\mid=\sqrt{a^2_x+a^2_y}$

$\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2$

Hence, the magnitude of acceleration when t=1 s= $64m/s^2$

3 0

2 years ago

What parts of an atom can change during a nuclear reaction that cannot change during a chemical reaction? Question 15 options: A

viva [34]

Neutrons and protons cannot be removed from nucleus from chemical reactions because they are strongly held together but nuclear reactions are strong enough to separate them. Hence option A is correct.

6 0

3 years ago

Read 2 more answers