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Misha Larkins [42]
3 years ago
12

Tin shears have longer handles than the scissors used to cut cloth why ​

Physics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

A pair of scissors used to cut a piece of cloth has blades longer than the handles so that the blades move longer on the cloth than the movement at the handles.

While shears used for cutting metals have short blades and long handles because as it enables us to overcome large resistive force by a small effort.

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Why do you need to breath faster at the top of a mountain
Dovator [93]
Because of the higher elevation. As much as we go away from earth by increasing altitude, there will occur lacoccuren. due to this, we need to breathe fast to get suffient oxygen by time at top of the mountain. We breathe faster at the top of a tall mountain due to a natural reflex. As we go higher up a mountain, the density of air and oxygen decreases, i.e. the amount of oxygen in the air decreases along with other gases. Therefore, to maintain the regular supply of oxygen, the body breathes faster. If we go too high without oxygen tanks, we could fall unconscious and die within 8 minutes due to oxygen deprivation.

Hope this helps! let me know if you need anything else!!

6 0
3 years ago
Which of the following is NOT an insulator?
sergejj [24]
I don’t know how air would be an insulator so I’m guessing that one isn’t
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3 years ago
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What are the similarities in the 3 types of waves
IgorLugansk [536]

Answer:

All kinds of waves have the same fundamental properties of reflection, refraction, diffraction, and interference, and all waves have a wavelength, frequency, speed, and amplitude.

Explanation:

7 0
3 years ago
A person throws a stone from the corner edge of a building. The stone's initial velocity is 28.0 m/s directed at 43.0° above the
Naya [18.7K]

The stone's acceleration, velocity, and position vectors at time t are

\mathbf a(t)=-g\,\mathbf j

\mathbf v(t)=v_{i,x}\,\mathbf i+\left(v_{i,y}-gt\right)\,\mathbf j

\mathbf r(t)=v_{i,x}t\,\mathbf i+\left(y_i+v_{i,y}t-\dfrac g2t^2\right)\,\mathbf j

where

g=9.80\dfrac{\rm m}{\mathrm s^2}

v_{i,x}=\left(28.0\dfrac{\rm m}{\rm s}\right)\cos43.0^\circ\approx20.478\dfrac{\rm m}{\rm s}

v_{i,y}=\left(28.0\dfrac{\rm m}{\rm s}\right)\sin43.0^\circ\approx19.096\dfrac{\rm m}{\rm s}

and y_i is the height of the building and initial height of the rock.

(a) After 6.1 s, the stone has a height of 5 m. Set the vertical component (\mathbf j) of the position vector to 5 m and solve for y_i:

5\,\mathrm m=y_i+\left(19.096\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.1\,\mathrm s)^2

\implies\boxed{y_i\approx70.8\,\mathrm m}

(b) Evaluate the horizontal component (\mathbf i) of the position vector when t=6.1\,\mathrm s:

\left(20.478\dfrac{\rm m}{\rm s}\right)(6.1\,\mathrm s)\approx\boxed{124.92\,\mathrm m}

(c) The rock's velocity vector has a constant horizontal component, so that

v_{f,x}=v_{i,x}\approx20.478\dfrac{\rm m}{\rm s}

where v_{f,x}

For the vertical component, recall the formula,

{v_{f,y}}^2-{v_{i,y}}^2=2a\Delta y

where v_{i,y} and v_{f,y} are the initial and final velocities, a is the acceleration, and \Delta y is the change in height.

When the rock hits the ground, it will have height y_f=0. It's thrown from a height of y_i, so \Delta y=-y_i. The rock is effectively in freefall, so a=-g. Solve for v_{f,y}:

{v_{f,y}}^2-\left(19.096\dfrac{\rm m}{\rm s}\right)^2=2(-g)(-124.92\,\mathrm m)

\implies v_{f,y}\approx-53.039\dfrac{\rm m}{\rm s}

(where we took the negative square root because we know that v_{f,y} points in the downward direction)

So at the moment the rock hits the ground, its velocity vector is

\mathbf v_f=\left(20.478\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(-53.039\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which has a magnitude of

\|\mathbf v_f\|=\sqrt{\left(20.478\dfrac{\rm m}{\rm s}\right)^2+\left(-53.039\dfrac{\rm m}{\rm s}\right)^2}\approx\boxed{56.855\dfrac{\rm m}{\rm s}}

(d) The acceleration vector stays constant throughout, so

\mathbf a(t)=\boxed{-g\,\mathbf j}

4 0
3 years ago
Please help! The image produced by a concave mirror is ? .
Alexeev081 [22]

Answer:

is a reflection.

The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted). When the object is that the focal point, the image is at infinity.

Explanation:

6 0
3 years ago
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