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CaHeK987 [17]
2 years ago
13

A tiger leaps with an initial velocity of 35.0 km/hr at an angle of 13.0ᶿ with respect to the horizontal. What are the component

s of the tiger’s velocity?
Physics
1 answer:
Lorico [155]2 years ago
3 0

Answer:

Vx = 35 x cos(13deg)

Vy = 35 x sin(13deg) - gt  

(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)

Explanation:

The tiger leaps up, then x and y component of its velocity are:

Vx = Vo x cos(alpha)

Vy = Vo x sin(alpha) - gt

(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)

Hope this helps!

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After an arrow is shot, is the force unbalanced or balanced? BRAINLY.
Reil [10]

Answer:

The force is unbalanced

Explanation:

After an arrow is shot, the force acting on the arrow is unbalanced. The resulting net force gives the arrow an initial acceleration which wanes with time and the body is brought to rest.

The net force acting on an arrow is not zero and this indicates that the forces acting on the arrow is unbalanced.

If the force is balanced, the arrow is expect to continue in uniform motion but that is not the case as air resistance has massive impact on this body.

7 0
2 years ago
Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J

Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
Which statement is not a good practice when working inside a computer case?
lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0
2 years ago
4. Dr. Copus is in charge of the cognition department at the University of Wisconsin-Madison. A new drug named Mem-Reen has beco
MrMuchimi

Answer:

See the answer below

Explanation:

<u>Independent variable</u>: Type of drug (Mem-Reen or placebo)

<u>Dependent variable</u>: memories

<u>Experimental group</u>: The group that was given Mem-Reen

<u>Control group</u>: The group that was given placebo

<u>Constants</u>: Food, hours of sleep, memory test procedures.

The independent variable is an input variable that produces effects on the dependent variable. As the variable is changed, it produces different effects on the dependent variable.

The dependent variable is the actual variable that is measured during an experiment. It is the main purpose of setting-up of an experiment.

The experimental group is also referred to as the treatment group while the control group is the group that does not receive treatment at all or they receive fake treatment/placebo.

Constants are unchanging variables included in experiments. They remain unchanged both in the treatment and the control group, otherwise, the outcome of the experiment will be unreliable.

5 0
2 years ago
The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time
Natali5045456 [20]
<span>x=((12.3/100)m)cos[(1.26s^−1)t]
 v= dx/dt = -</span><span>((12.3/100)*1.26)sin[(1.26s^−1)t]
 v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
 v=<span> <span>-0.13261622 m/s
 </span></span>the object moving at  0.13 m/s <span>at time t=0.815 s</span>
6 0
2 years ago
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