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Lelu [443]
2 years ago
13

Out of hypothesis, theories and laws “can help predict future events”

Physics
1 answer:
Amanda [17]2 years ago
7 0

Answer:

true

Explanation:

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A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H
Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

\theta =44.8{\circ}

using v^2-u^2=2as

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration   (gsin\theta )

s=distance traveled

0-(1.72)^2=2(-9.81\times sin(44.8))s

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

0=1.72-gsin(44.8)\times t

t=\frac{1.72}{9.8\times sin(44.8)}

t=0.249 s

(c)Speed of the block at bottom

v^2-u^2=2as

Here u=0 as it started coming downward

v^2=2\times gsin(44.8)\times 0.214

v=\sqrt{2.985}

v=1.72 m/s

3 0
3 years ago
A helicopter is hovering at a constant height of 35 m. The upward lift force on the helicopter is 85500. What is the weight of t
Vika [28.1K]
2,442! I think..right ok
7 0
2 years ago
What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?
ch4aika [34]

Answer:

c=10\ J/kg^{\circ} C

Explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature, \Delta T=6^{\circ} C

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C

So, the specific heat of the object is 10\ J/kg^{\circ} C.

5 0
2 years ago
A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe
schepotkina [342]

Answer:

Explanation:

Given that,

Mass of block

M = 2kg

Spring constant k = 300N/m

Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

Vo = 12m/s

Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

k is spring constant = 300N/m

m is the mass of object = 2kg

Then,

ω = √300/2 = √150

ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

A = √[0 + (12²/12.5²)]

A = √[0 + 0.96]

A = √0.96

A = 0.98m

4 0
2 years ago
A scientist that applies the laws of science to the needs of communities is called _____.
s2008m [1.1K]

Answer:

The experimental scientist

5 0
2 years ago
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