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marin [14]
2 years ago
7

An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of

the object makes an angle α0 with the horizontal. Neglect air resistance. If α0 is 90∘, so that the object is thrown straight up (but misses the roof on the way down), what is the speed v of the object just before it strikes the ground?
Physics
1 answer:
Vsevolod [243]2 years ago
6 0

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

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A 3.50-meter length of wire with a cross-sectional area of 3.14 × 10-6 meter2 is at 20° Celsius. If the wire has a resistance of
maks197457 [2]

Answer:

5.6\times 10^{-8}\ Ohm.m

Explanation:

Resistivity is given by \rho=\frac {AR}{L} where A is cross-sectional area, R is resistance, L is the length and \rho is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to

\rho=\frac {3.14\times 10^{-6}\times 0.0625}{3.5}=5.60714285714285714285714285714285714285\times 10^{-8}\approx 5.6\times 10^{-8}\ Ohm.m

8 0
2 years ago
A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t
alexira [117]

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

5 0
1 year ago
A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe
galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m,  y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}

The velocity v is given by:

\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}

The acceleration a:

\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}

From the given values we get two equations:

-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4

We also know:

\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}

The magnitude of the acceleration a is:

a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7

The magnitude of position r is:

r=R=68m

Plugging in to the equation for a(t):

\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}

and solving for ω:

|\omega|=0.66

Now solve for time t:

\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83

Using the calculated values to compute v(t):

\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}

The speed of the car is:

\sqrt{38.3^2 + (-23.5)^2} = 44.9

The position r:

\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}

5 0
2 years ago
Read 2 more answers
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
ycow [4]
The mass of the puck is
m = 0.15 kg.
The diameter of the puck is 0.076 m, therefore its radius  is
r = 0.076/2 = 0.038 m
The sliding speed is
v = 0.5 m/s
The angular velocity is
ω = 8.4 rad/s

The rotational moment of inertia of the puck is
I = (mr²)/2
  = 0.5*(0.15 kg)*(0.038 m)²
  = 1.083 x 10⁻⁴ kg-m²

The kinetic energy of the puck is the sum of the translational and rotational kinetic energy.
The translational KE is
KE₁ = (1/2)*m*v²
       = 0.5*(0.15 kg)*(0.5 m/s)²
       = 0.0187 j

The rotational KE is
KE₂ = (1/2)*I*ω²
       = 0.5*(1.083 x 10⁻⁴ kg-m²)*(8.4 rad/s)²
       = 0.0038 J

The total KE is
KE = 0.0187 + 0.0038 = 0.0226 J

Answer: 0.0226 J


4 0
1 year ago
Researchers interested in studying stress gave 150 high school seniors a very difficult math exam. After the test, the researche
belka [17]

Answer:

<u>Informed Consent- The study didn't conform this ethical standard.</u>

<u>Debriefing- The study conforms this ethical standard.</u>

<u>Confidentiality- The study conforms this ethical standard.</u>

<u>Protection from harm- The study conforms this ethical standard.</u>

Explanation:

<u>Informed consent</u> is described as a procedure whereby researchers tend to provide details about specific research they are going to conduct, the risk & benefits involved in the study, different alternatives involved in the procedure, etc.

<u>Debriefing </u>is described as a process that is being conducted in any of the different psychological research encompassing human participants after specific research is completed. The researcher tends to describe the details of the research to the participants.

<u>Confidentiality</u> is described as one of the different code of ethics followed by health workers or psychologists. While practicing confidentiality, a psychologist tends to promise his or her participants that he or she will keep everything a secret whatever is being discussed or shared between both of them.

<u>Protection from harm</u> determines that the psychologists conducting research follows the ethics in which he or she has to protect all the participants from any kind of harm.

4 0
1 year ago
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