What two products are formed from the fission of uranium?

A 2.0 kg block, initially moving at 10.0 m/s, slides 50.0 m across a sheet of ice beforecoming to rest. What is the magnitude of

inna [77]

Answer:

The magnitude of the average frictional force on the block is 2 N.

Explanation:

Given that.

Mass of the block, m = 2 kg

Initial velocity of the block, u = 10 m/s

Distance, d = 50 m

Finally, it stops, v = 0

Let a is the acceleration of the block. It can be calculated using third equation of motion. It can be given by :

$v^2-u^2=2ad$

$-u^2=2ad$

$a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(10\ m/s)^2}{2\times 50\ m}\\\\a=-1\ m/s^2$

The frictional force on the block is given by the formula as :

F = ma

$F=2\ kg\times (-1)\ m/s^2\\\\F=-2\ N$

|F| = 2 N

So, the magnitude of the average frictional force on the block is 2 N. Hence, this is the required solution.

8 0

3 years ago

Worth 25 points on my exam

I am Lyosha [343]

Initial velocity=20m/s
Final velocity=0m/s(As the car stops)
Acceleration=-8m/s^2
Distance=s=26m

We need to verify the thrid equation of kinematics here

$\\ \tt\longmapsto v^2-u^2=2as$

$\\ \tt\longmapsto 20^2=2(-8)s$

$\\ \tt\longmapsto 400=-16s$

$\\ \tt\longmapsto s=|400/-16|$

$\\ \tt\longmapsto s=25m$

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

7 0

2 years ago

Read 2 more answers

A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the rail

disa [49]

Answer

given,

mass of ice hockey player = 110 Kg

initial speed of the skate = 3 m/s

final speed of the skate = 0 m/s

distance of the center of mass, m = 30 cm = 0.3 m

a) Change in kinetic energy

$\Delta KE = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2$

$\Delta KE = \dfrac{1}{2}m(0)^2 - \dfrac{1}{2}\times 110 \times 3^2$

$\Delta KE = - 495\ J$

b) Average force must he exerted on the railing

using work energy theorem

W = Δ KE

F .d = -495

F x 0.3 = -495

F = -1650 N

the average force exerted on the railing is equal to 1650 N.

7 0

3 years ago

Read 2 more answers

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo

makkiz [27]

Answer:

(a) The speed of the target proton after the collision is: $V_{2f} =433(m/s)$ , and (b) the speed of the projectile proton after the collision is: $v_{1f}=250(m/s)$ .

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle: $m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}$ , and y axle: $0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}$ . Now replacing the value given as: $v_{1i}=500(m/s)$ , $\beta_{1}=+60^{o}$ for the projectile proton and according to the problem $\beta_{1}and\beta_{2}$ are perpendicular so $\beta_{2}=-30^{o}$ , and assuming that $m_{1}=m_{2}$ , we get for x axle: $500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2}$ and y axle: $0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}$ , then solving for $v_{2f}$ , we get: $v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f}$ and replacing at the first equation we get: $500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}$ , now solving for $v_{1f}$ , we can find the speed of the projectile proton after the collision as: $v_{1f}=250(m/s)$ and $v_{2f}=\sqrt{3}*v_{1f}=433(m/s)$ , that is the speed of the target proton after the collision.

5 0

3 years ago

A kid pulls on a rope with 20 newtons of force. The block and tackle system pulls up a 160 newton box. What is the mechanical ad

Minchanka [31]

Answer:
The mechanical advantage of the system is 8

Explanation:
the mechanical advantage measures how much the system multiplies the input force to get the output.

In the given:
The input force (effort) is 20 Newton
The output force (load) is 160 Newton

This means that the mechanical advantage is:
mechanical advantage = load / effort = 160 / 20 = 8

Note that the mechanical advantage is unit-less (has no unit) since it is a ratio between two forces.

Hope this helps :)

3 0

3 years ago