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3 years ago

Measurements made by a caliper can be read from a dial gage, a digital display, or a

1 answer:

7 0

<span> vernier scale because is a part of vernier calipers.rest of them are parts of other measuring instruments like dial guage, screw guage etc</span>

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Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30 m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let $\theta$ be the angle .

Therefore,

$30 cos\theta=12.6$

$cos\theta=\frac{12.6}{30}=0.42$

$\theta=cos^{-1}(0.42)=65.165^{\circ}$

b.We have to find the height to which ball reach.

$v^2-v^2_0=2aS$

$S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}$

$S=37.78 m$

7 0

3 years ago

The statement “Heavy objects fall faster than light objects” is an example of a(n) _______.

vazorg [7]

D. Theory is the answer

7 0

3 years ago

An artillery shell is launched on a flat, horizontal field at an angle of α = 40.8° with respect to the horizontal and with an i

Lina20 [59]

Answer:

1317.4 m

Explanation:

We are given that

Angle= $\alpha=40.8^{\circ}$

Initial speed = $v_0=346m/s$

We have to find the horizontal distance covered by the shell after 5.03 s.

Horizontal component of initial speed= $v_{ox}=v_0cos\theta=346cos40.8=261.9m/s$

Vertical component of initial speed= $v_{oy}=346sin40.8=226.1m/s$

Time=t=5.03 s

Horizontal distance = $Horizontal\;velocity\times time$

Using the formula

Horizontal distance= $261.9\times 5.03$

Horizontal distance=1317.4 m

Hence, the horizontal distance covered by the shell=1317.4 m

8 0

3 years ago

Why do you think the standards changed over time?

Eddi Din [679]

Answer:

https://www.quora.com/Why-do-you-think-standards-of-beauty-have-changed-over-the-years

Explanation:

your answer is in this link

7 0

3 years ago

The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.

klio [65]

The relationship between frequency and wavelength for an electromagnetic wave is
$c=f \lambda$
where
f is the frequency
$\lambda$ is the wavelength
$c=3 \cdot 10^8 m/s$ is the speed of light.

For the light in our problem, the frequency is $f=1.20 \cdot 10^{13} s^{-1}$ , so its wavelength is (re-arranging the previous formula)
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}= 2.5 \cdot 10^{-5}m$

8 0

3 years ago