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ira [324]
3 years ago
6

Less force is required to poke through an eggshell from the _____.

Physics
1 answer:
meriva3 years ago
8 0
I believe the answer is A. hope this helps :)
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Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as
viva [34]

Answer:

a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s

Explanation:

It is angular momentum given by

      L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]

Let's apply this relationship to our case

Let's start by breaking down the speed

      v₀ₓ = v₀ cosn 45

      voy =v₀ sin 45

      v₀ₓ = 9 cos 45

      voy = 9 without 45

      v₀ₓ = 6.36 m / s

      voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

   vfy² = voy²- 2 g y

   y = voy² / 2g

   y = (6.36)²/2 9.8

   y = 2.06 m

Let's calculate the angular momentum

L= \left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^   Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

     R = vo² sin 2θ / g

     R = 9² sin (2 45) /9.8

     R = 8.26 m

L = \left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^   kg m² /s

5 0
3 years ago
Why do you think the standards changed over time?
Eddi Din [679]

Answer:

https://www.quora.com/Why-do-you-think-standards-of-beauty-have-changed-over-the-years

Explanation:

your answer is in this link

7 0
3 years ago
An important thing to consider when responding to a driver in front of you that stops suddenly is:a. the mental state of the oth
Gre4nikov [31]
The correct answer is option A. i.e. An important thing to consider when responding to a driver in front of you that stops suddenly is: the mental state of the other driver.

Our talk or discussion can disrturb the balance of the driver or he can get distracted. So, we must try not to speak much while the driver is driving because by doing this we are putting the life of ourselves in danger. Any distrcaction of driver can cause accident.
5 0
3 years ago
A diver jumps off a diving platform that is 20 meters long. Describe the transfer of energy that occurs during the fall.
kobusy [5.1K]

Answer: gravitational potential energy is converted into kinetic energy

Explanation:

When the diver stands on the platform, at 20 m above the surface of the water, he has some gravitational potential energy, which is given by

E=mgh

where m is the man's mass, g is the gravitational acceleration and h is the height above the water. As he jumps, the gravitational potential energy starts decreasing, because its height h above the water decreases, and he acquires kinetic energy, which is given by

K=\frac{1}{2}mv^2

where v is the speed of the diver, which is increasing. When he touches the water, all the initial gravitational potential energy has been converted into kinetic energy.

8 0
3 years ago
A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh
daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is B=2.958\times10^{-5}T

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \  \overrightarrow B_s\perp \overrightarrow B_w

Angle of net magnetic field from axial direction is given by:

tan\  \theta=\frac{B_w}{B_s},

Field due to solenoid:

B_s=\mu_onI_s,  \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m

Field due to wire:

B_w=\frac{\mu_oI_w}{2\pi r}

Therefore, r:

tan\  \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi  nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T

Hence, the magnetic field is B=2.958\times10^{-5}T

7 0
3 years ago
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