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hjlf
3 years ago
12

HELP PLEASE ITS URGENT WORK DUE TOMORROW!!!!!!!!!!!

Physics
1 answer:
Gemiola [76]3 years ago
7 0

Answer:

Just dont it but shhh dont tell your parent about this

Explanation:Just dont do but shhh dont tell your parent about thisJust dont do but shhh dont tell your parent about this

Just dont do but shhh dont tell your parent about thisJust dont do but shhh dont tell your parent about thisJust dont do but shhh dont tell your parent about thisJust dont do bJust dont do but shhh dont tell your parent about thisut shhh dont tell your parent about this

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A bullet of mass 6.20 10-3 kg, moving at 1320 m/s impacts a tree stump and penetrates 11.00 cm into the wood before coming to re
sladkih [1.3K]

Answer:

  F = -49.1   10³ N

Explanation:

Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant

   v_{f}² = v₀² + 2 a x

Since the bullet is at rest, the final speed is zero

    x = 11.00 cm (1 m / 100 cm) = 0.110 m

    0 = v₀² + 2 a x

   a = -v₀² / 2 x

   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

   F = m a

   F = 6.20 10⁻³ (-7.92 10⁶)

   F = -49.1   10³ N

The sign means that it is the force that the tree exerts to stop the   bullet

8 0
3 years ago
A horse was grazing at the top of a small hill. The horse got thirsty and ran toward a
Phoenix [80]

Answer:

I think it may be decreased, but i'm in 6th grade so-

Explanation:

7 0
3 years ago
Potential energy results from the ______or position of an object.
Burka [1]

Answer:

Mass

Explanation:

8 0
3 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
Which of the following statements about electric field lines associated with electric charges is false? Electric field lines can
Tcecarenko [31]

Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

  • Electric field lines originate from positive end and terminates at negative end,i.e., field lines are inward in direction to the negative charges and outward from the positive charges.
  • These lines when close together represents high intensity and when far apart shows low intensity of the field.
  • These lines do not intersect, as the tangent drawn on these lines provides us with the field direction and intersection of these lines means two field directions which is not possible.
  • These lines unlike magnetic field lines do not form closed loops as they do not turn around but originate at positive end and terminates at negative end which ensures no loop formation.
8 0
4 years ago
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