Which of the following is true of our current knowledge of electrons?

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago

Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released)

cluponka [151]

<span>Exothermic reaction is a chemical reaction that releases energy. This reaction releases heat energy or light . In an endotermic reaction energy is used. Enthaply is the heat energy change , delta H. If the sum of the enthalpies of the reactans is greater than the products the reaction is exothermic. If the products side has a larger enthaply than the process is endothermic. So, if delta H is negative then the process is exothermic. If delta H is positive, than the process is endothermic. Exothermic are: A+BC -> AB+C A2+B2 -> 2AB Endothermic are:AB+C -> AC+B A2 + C2 -> 2AC B2+C2 -> 2BC</span>

5 0

3 years ago

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.

Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

$KE_i+PE_i =KE_f+PE_f$

or

$\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh$

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

$u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s$

4 0

3 years ago

The angle between the axes of two polarizing filters is 45.0^\circ45.0 ∘ . By how much does the second filter reduce the int

suter [353]

Answer

given,

angle between two polarizing filters = 45°

filter reduce intensity = ?

a) I = I₀ Cos² θ

here θ = 45⁰

$I = \dfrac{I_0}{2}$

intensity of the light is reduced by 0.500

correct answer from the given option D

b) direction of the polarization

θ = 45°

7 0

3 years ago