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damaskus [11]
3 years ago
5

Help please!!!

Physics
1 answer:
TiliK225 [7]3 years ago
7 0

Answer:

Cu 8.92

Explanation:

The formula for density is mass/volume. If you were to divide 62.44 by 7, you would get 8.92. Since copper is the only metal in this table that has a density of 8.92, that is the answer.

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A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the
brilliants [131]

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

y = yo+ V_y t +\frac{1}{2}gt^2

0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2

solving for t

t = 1.075 sec

for horizontal motion

x = V_x t

x = 2cos18*1.075

x = 2.044 m

8 0
3 years ago
Distinguish
yaroslaw [1]

Answer:

Velocity and speed both are continuously increasing.

Acceleration is constant.

Explanation:

Speed is defined as length of path covered by a body per unit time. Speed is a scalar quantity that consist of magnitude only and not direction.

Velocity is defined as the displacement per unit times. Displacement is the shortest distance between the two points. It is a vector quantity and  hence has a direction in the direction of displacement along with its own magnitude.

  • Both velocity and speed have same unit of measure which is meter per second in S.I. During <em>free fall</em> in the absence of any air resistance the velocity and speed both will be having a vertical downward direction with continuously increasing magnitude. Tough we are not concerned about the direction when discussing about speed but here both are equal since the motion is linear.

Acceleration is the rate of change in velocity of a body which is a vector quantity. For speed we are concerned about instantaneous acceleration since for a short period of time it may have a specific direction.

  • During free fall the acceleration is of a body is equal to the acceleration due to gravity and constant when the height of fall is much lesser than the radius of the earth.
4 0
3 years ago
Activity 1. Part A. Lat There Be Light
Pani-rosa [81]
I thinks it’s 2...........
6 0
2 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t
Sladkaya [172]

Answer:

0.767

Explanation:

The work done on the truck by the frictional drag force is given by

W=-Fd

where

F is the magnitude of the frictional force

d = 38.0 m is the maximum displacement allowed for the truck

The negative sign is due to the fact that the force of friction is opposite to the motion of the truck

The force of friction can also be written as:

F=\mu mg

where

\mu is the coefficient of kinetic friction between the truck and the lane

m is the mass of the truck

g is the acceleration of gravity

So we can rewrite the work done as

W=-\mu mg d (1)

According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the truck:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2 (2)

where

v = 0 is the final velocity of the truck

u = 23.9 m/s is the initial velocity of the truck

By combining (1) and (2) we get

-\frac{1}{2}mu^2 = -\mu mg d

And solving for \mu, we find the minimum coefficient of kinetic friction able to stop the truck in a distance d:

\mu = \frac{u^2}{2gd}=\frac{23.9^2}{2(9.8)(38.0)}=0.767

7 0
3 years ago
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