1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ludmilkaskok [199]
3 years ago
9

If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the magnitude of the resultant velocity?

Physics
1 answer:
kramer3 years ago
7 0

Answer:

40 m/s

Explanation:

You might be interested in
The height of the Washington Monument is measured to be 170.00 m on a day when the temperature is 35.0°C. What will its height b
N76 [4]

Answer:

169.98 m

Explanation:

\alpha = Thermal coefficient of marble = 2.5\times 10^{-6}/^{\circ}C

L_0 = Initial length = 170 m

\Delta T = Change in temperature = (-10-35)°C

Change in length is given by

\Delta L=\alpha L_0\Delta T\\\Rightarrow \Delta L=2.5\times 10^{-6}\times 170\times (-10-35)\\\Rightarrow \Delta L=-0.019125\ m

The height on the given day will be 170-0.019125=169.98\ m

3 0
4 years ago
A 2.0-kg silverware drawer does not slide readily. The owner gradually pulls with more and more force, and when the applied forc
topjm [15]

Answer: 0.45

Explanation:

First note that the body that causes the body to move is its moving force (Fm) which is 9.0N

Since the mass of the body is 2.0kg, the weight will be;

W= mg = 2×10

W= 20N

For static body, the frictional force (Ff) acting on the body is equal to the moving force (Fm) since both forces acts along the horizontal on the body.

Ff = Fm = 9.0N

The normal reaction (R) on the body will also be equal to its weight(W) since weight acts downwards and the reaction acts in the opposite direction (upwards).

R = W = 20N

Ff = nR taking 'n' as coefficient of static friction between the drawer and the cabinet.

9.0 = 20n

n = 9/20

n = 0.45

7 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:

a)

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.

b)

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

c)

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

d)

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.

e)

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:

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
4 years ago
Kepler's laws, satellites motion and weightlessness
Anon25 [30]

Answer:

First Kepler law states that <em><u>Each</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>describes</u></em><em><u> </u></em><em><u>an</u></em><em><u> </u></em><em><u>ellipsoidal</u></em><em><u> </u></em><em><u>motion</u></em><em><u> </u></em><em><u>about</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>its</u></em><em><u> </u></em><em><u>single</u></em><em><u> </u></em><em><u>focus</u></em><em><u>.</u></em>

Second Kepler law states that <em><u>A</u></em><em><u>n</u></em><em><u> </u></em><em><u>i</u></em><em><u>m</u></em><em><u>a</u></em><em><u>g</u></em><em><u>i</u></em><em><u>n</u></em><em><u>a</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>l</u></em><em><u>i</u></em><em><u>n</u></em><em><u>e</u></em><em><u> </u></em><em><u>j</u></em><em><u>o</u></em><em><u>i</u></em><em><u>n</u></em><em><u>i</u></em><em><u>n</u></em><em><u>g</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>t</u></em><em><u>o</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>Sun</u></em><em><u> </u></em><em><u>sweeps</u></em><em><u> </u></em><em><u>out</u></em><em><u> </u></em><em><u>equal</u></em><em><u> </u></em><em><u>areas</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>equal</u></em><em><u> </u></em><em><u>time</u></em><em><u> </u></em><em><u>intervals</u></em><em><u>.</u></em>

Third Kepler law states that <em><u>The</u></em><em><u> </u></em><em><u>squares</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>period</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>revolution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>around</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>proportional</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>cubes</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>mean</u></em><em><u> </u></em><em><u>distance</u></em><em><u> </u></em><em><u>between</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u>.</u></em>

Weightlessness is the condition where the body has zero gravity ( its acceleration is equal to the acceleration due to gravity )

.

3 0
3 years ago
The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re
weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

5 0
3 years ago
Other questions:
  • HELP PLEASE
    15·1 answer
  • Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca
    8·1 answer
  • Find the current drawn from the battery by the network of four resistors shown in the figure.
    15·1 answer
  • A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cyc
    15·1 answer
  • A strong lightning bolt transfers about 25 c to earth. how many electrons are transferred? the elementary charge is 1.60218 × 10
    8·1 answer
  • Troposphere is the highest atmospheric layer.
    12·1 answer
  • At what temperature the semiconductor behaves like conductor?​
    9·1 answer
  • URGENT! PLEASE! how might the speed of an electromagnetic wave be affected as it moves from a gas (such as the atmosphere) to a
    10·1 answer
  • The slit of width ‘d’ is illuminated by light of wavelength 5000a0. For what value of ‘d’ will the first maximum fall at an angl
    11·1 answer
  • Tom travels to the moon for NASA. While on the moon, the total weight of Tom and his lunar exploration suit is 224 N. His suit h
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!