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3 years ago

What type of object is called the object at rest

1 answer:

8 0

Rest is the state of an object being stationary relative to a particular frame of reference or another object; when the position of a body with respect to its surroundings does not change with time it is said to be "at rest".

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A projectile is launched at ground level with an initial speed of 50.0m/s at an angle of 30° above the horizontal. It strikes a

tino4ka555 [31]

Answer:

x = 129.9 m

y = 30.9 m

Explanation:

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

$v_{i}$ = 50 m/s

Angle = 30°

Time = t = 3 s

horizontal component of velocity = $v_{i_{x}}$ = $v_{i}$ cos30°

$v_{i_{x}}$ = 50cos30°

$v_{i_{x}}$ = 43.3 m/s

Vertical component of velocity = $v_{i_{y}}$ = $v_{i}$ Sin30°

$v_{i_{y}}$ = 50Sin30°

$v_{i_{y}}$ = 25 m/s

This is a projectile motion, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

But the vertical component of velocity varies with time and there is an acceleration along vertical direction which is equal to gravitational acceleration g.

Horizontal distance = x = $v_{i_{x}}$ t

x = 43.3*3

x = 129.9 m

Vertical Distance = y = $v_{i_{y}}$ t -0.5gt²

y = 25*3 - 0.5*9.8*3²

y = 75 - 44.1

y = 30.9 m

3 0

3 years ago

20- A gram of distilled water at 4° C:

Usimov [2.4K]

Answer:

D. will decrease slightly in volume when heated to 6° C

Explanation:

6 0

3 years ago

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of 1.0 nC/mm2 . If the potential between t

Archy [21]

Answer:

$A=81mm^2$

Explanation:

We know that for a capacitor $Q=CV$ , where Q is the charge of one plate, C the capacitance and V the potential between the plates.

We also know that $Q=\sigma A$ , since $\sigma$ is the surface charge density and A the area of the plate (both equal in our case).

Putting all together:

$A=\frac{CV}{\sigma}$

Which for our values is:

$A=\frac{(3\times10^{-12}F)(27\times10^3V)}{1\times10^{-9}C/mm^2}=81mm^2$

Where we notice that the S.I. units combination FV/C must not have units (we can verify it directly from their definitions or we notice that $mm^2$ is enough to describe an area).

8 0

3 years ago

An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.4m/s from a height of 1.55m above th

Natali5045456 [20]

a) Yes, the rock will reach the top

b) The final speed is 3.7 m/s

c) The change in speed is 2.4 m/s

d) The change in speed in the two situations do not agree

e) Because the kinetic energy depends quadratically on the speed, $K\propto v^2$

Explanation:

The mechanical energy of the rock at the moment it is thrown from the ground is equal to the sum of its kinetic energy and its potential energy:

$E=KE_i + PE_i = \frac{1}{2}mu^2 + mgh_i$

where

m is the mass of the rock

u = 7.4 m/s is the inital speed

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 1.55 m$ is the initial height of the rock

Substituting, we find the initial mechanical energy of the rock

$E=\frac{1}{2}m(7.4)^2 + m(9.8)(1.55)=42.6m$ [J]

In order to reach the top of the castle, the rock should have a mechanical energy of at least

$E' = mgh'$

where

h' = 3.65 m is the heigth of the top

Substituting,

$E'=m(9.8)(3.65)=35.6m$ [J]

Since E > E', it means that the rock has enough mechanical energy to reach the top.

The final mechanical energy of the rock at the top is

$E=mgh'+ \frac{1}{2}mv^2$ (1)

where:

v is the final speed of the rock at the top

Since the mechanical energy is conserved, this should be equal to the initial mechanical energy:

$E=42.6 m$ [J] (2)

Therefore, equating (1) and (2), we can find the final speed of the rock:

$mgh' + \frac{1}{2}mv^2 = 42.6m\\v=\sqrt{2(42.6-gh')}=\sqrt{2(42.6-(9.8)(3.65))}=3.7 m/s$

Since the motion of the rock is a free fall motion (constant acceleration equal to the acceleration of gravity), we can use the following suvat equation:

$v^2 - u^2 = 2as$

where

v is the final speed, at the bottom

u = 7.4 m/s is the initial speed of the rock, at the top

$a=9.8 m/s^2$ is the acceleration of gravity

s = 3.65 - 1.55 = 2.1 m is the vertical displacement of the rock

Solving for v, we find the final speed:

$v=\sqrt{u^2+2as}=\sqrt{7.4^2 + 2(9.8)(2.1)}=9.8 m/s$

Therefore, the change in speed is

$\Delta v = v-u = 9.8 - 7.4 =2.4 m/s$

In the first situation (rock thrown upward), we have:

u = 7.4 m/s (initial speed)

v = 3.7 m/s (final speed)

So the change in speed is

$\Delta v = v-u =3.7 - 7.4 = -3.7 m/s$

While the change in speed in the second situation (rock thrown downward) is

$\Delta v = 2.4 m/s$

Therefore, we see that their magnitudes do not agree.

In both situations, the change in kinetic energy of the rock is equal in magnitude to the change in gravitational potential energy, since the total mechanical energy is conserved.

The change in gravitational potential energy in the two situations is the same (because the change in height is the same), therefore the change in kinetic energy in the two situations is also the same.

However, the kinetic energy of the rock is not directly proportional to the speed, but to the square of the speed:

$K\propto v^2$

Since the initial speed is the same for both situation (7.4 m/s), but the change in kinetic energy has opposite sign in the two situations (negative when the rock is thrown upward, positive when thrown downward), the situation is not symmetrical, therefore in order to have the same magnitude of change in the kinetic energy, the change in speed must be larger when the kinetic energy involved is lower, so in the first situation.

Learn more about kinetic energy and about potential energy:

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#LearnwithBrainly

6 0

4 years ago

A girl walks 5km due north,then in the direction of 60° of northeast towards her final destination.If the magnitude of her resul

storchak [24]

Answer:

4.2 km

Explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as isosceles triangle, scalene triangle, equilateral triangle and right angled triangle.

Sine rule states that for a triangle with angles A,B and C and corresponding opposite sides a, b, and c, the following formula holds:

$\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$

Let the starting point be A, hence the side opposite to the angle = a = distance travelled in north east direction.

Let the point of 60° of northeast be B = 90° + (90° - 60°) = 120°, hence the side opposite to the angle = b = resultant displacement = 8 km.

Let C be the endpoint, hence the side opposite to the angle = c = distance travelled north = 5 km

Using sine rule:

$\frac{b}{sin(B)}=\frac{c}{sin(C)} \\\\\frac{8}{sin(120)}=\frac{5}{sin(C)} \\\\sin(C)=\frac{5*sin(120)}{8} =0.5412\\\\C=sin^{-1}(0.5412)=32.8^o\\\\$

∠A + ∠B + ∠C = 180° (sum of angle in a triangle)

∠A + 120 + 32.8 = 180

∠A = 27.2°

$\frac{b}{sin(B)}=\frac{a}{sin(A)} \\\\\frac{8}{sin(120)}=\frac{a}{sin(27.2)} \\\\a=\frac{8*sin(27.2)}{sin(120)} \\\\a=4.2\ km$

Therefore she travelled 4.2 km in the north east direction

5 0

3 years ago

What type of object is called the object at rest​

What type of object is called the object at rest