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Greeley [361]
3 years ago
13

A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an

d the mass of the sled is 10.3 kg.
The force of friction acting on the sled as it slides is -123 N.

If the sled was initially moving at 13.3 m/s at t=0, how far will it slide before the sled stops?
Physics
1 answer:
jonny [76]3 years ago
6 0

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

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Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
kirza4 [7]

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

6 0
3 years ago
When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it fr
Scrat [10]

Answer:

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

7 0
3 years ago
The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on
Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0
3 years ago
How many neutrons are needed to initiate the fission reaction shown?
nalin [4]

Answer: One neutron

Explanation:

one neutron 1/0n

Sum up the mass numbers on the right 99 + 135 + 2 = 236.

The sum of the mass numbers on the left should equal 236. 235 + 1 = 236

4 0
3 years ago
A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s
Alisiya [41]

Answer:

v_{ic}=92.53 m/s

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

p_{i}=p_{f}

m_{c}v_{ic}=(m_{c}+m_{w})V (1)

  • m(c) is the mass of stick clay
  • m(w) is the mass of the wooden block
  • v(ic) is the initial velocity of clay
  • V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

\Delta E=W

\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d

\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd

We can find V from this equation:

V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s

Now, let's put V into the equation (1) and find v(ic)

v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s

I hope it helps you!  

<u />

8 0
2 years ago
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