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4 years ago

Where is there potential energy throughout the loading, cocking, and releasing if the trebuchet?

Physics

1 answer:

pshichka [43]4 years ago

4 0

Answer:

adapted from NOVA, a team of historians, engineers, and trade experts recreate a medieval throwing machine called a trebuchet. To launch a projectile, a trebuchet utilizes the transfer of gravitational potential energy into kinetic energy. A massive counterweight at one end of a lever falls because of gravity, causing the other end of the lever to rise and release a projectile from a sling. As part of their design process, the engineers use models to help evaluate how well their designs will work.

Explanation:

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It's always a good idea to wear a seatbelt in a car, because if the car comes to a sudden stop, you will not _________.

IgorLugansk [536]

It's always a good idea to wear a seatbelt in a car, because if the car comes to a sudden stop, you will not move forward very much since the seatbelt is holding you back. The answer is letter D.

4 0

4 years ago

Barnard's star is a near neighbor of the Sun whose properties we know quite well. It is a type M4V with absolute magnitude 13.22

vitfil [10]

Answer:

The star is at a distance of 100 parsecs.

Explanation:

The distance can be determined by means of the distance modulus:

$M - m = 5log(d) - 5$ (1)

Where M is the absolute magnitude, m is the apparent magnitude and d is the distance in units of parsec.

Therefore, d can be isolated from equation 1

$log(d) = (M - m + 5)/5$

Then, Applying logarithmic properties it is gotten:

$d = 10^{(M - m + 5)/5}$ (2)

The absolute magnitude is the intrinsic brightness of a star, while the apparent magnitude is the apparent brightness that a star will appear to have as is seen from the Earth.

Since both have the same spectral type is absolute magnitude will be the same.

Finally, equation 2 can be used:

$d = 10^{(13.22 - 8.22+ 5)/5}$

$d = 100 pc$

Hence, the star is at a distance of 100 parsecs.

Key term:

Parsec: Parallax of arc seconds

8 0

3 years ago

Lightning As a crude model for lightning, consider the ground to be one plate of a parallel-plate capacitor and a cloud at an al

olga nikolaevna [1]

Answer:

$6.67154\times 10^{-9}\ F$

13.009503 C

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

k = Dielectric constant of air $3\times 10^6\ V/m$

Side of plate = 0.7 km

A = Area

d = Distance = 650 m

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 700^2}{650}\\\Rightarrow C=6.67154\times 10^{-9}\ F$

The capacitance is $6.67154\times 10^{-9}\ F$

Electric field is given by

$Q=CV\\\Rightarrow Q=Ckd\\\Rightarrow Q=6.67154\times 10^{-9}\times 3\times 10^6\times 650\\\Rightarrow Q=13.009503\ C$

The charge on the cloud is 13.009503 C

7 0

3 years ago

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

a runner moves 2.88 m/s north. she accelerates at 0.350 m/s2 at -52.0 angle. at the point where she is running directly east, wh

Naily [24]

Answer:

11.7 m

Explanation:

I assume north is the y direction and x is the east direction, so Δx refers to the displacement in the east direction.

First, find the time it takes for the velocity to change from directly north to directly east.

Given (in the y direction):

v₀ = 2.88 m/s

v = 0 m/s

a = 0.350 m/s² sin(-52.0°) = -0.276 m/s²

Find: t

v = at + v₀

(0 m/s) = (-0.276 m/s²) t + (2.88 m/s)

t = 10.4 s

Given (in the x direction):

v₀ = 0 m/s

a = 0.350 m/s² cos(-52.0°) = 0.215 m/s²

t = 10.4 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10.4 s) + ½ (0.215 m/s²) (10.4 s)²

Δx = 11.7 m

7 0

3 years ago