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ss7ja [257]
3 years ago
8

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

at is the friction force you must overcome to push the cart up the ramp
Physics
1 answer:
pentagon [3]3 years ago
4 0

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

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A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the groun
zheka24 [161]

Answer:

a) t = 2.64s

b) Vf = -28.7m/s

Explanation:

If the balloon is descending, the velocity is -2.3m/s. So the equation to describe the postion of the falling camera is:

Y = Vo*t - \frac{g*t^{2}}{2}

-41 = -2.3*t - \frac{10*t^{2}}{2}    Solving for t, we get:

t1 = -3.1s     and      t2 = 2.64s We discard the negative time and use the positive one.

The velocity of the camera will be:

Vf = Vo - g*t = -2.3 - 10*2.64 = -28.7m/s

6 0
3 years ago
PLEASE ASAP ILL GIVE BRAINLIEST.
taurus [48]

Answer:

applied force

Explanation:

any force where you push or pull is always applied force.

4 0
3 years ago
Please show your work.<br> 13. Convert 97 degrees fahrenheit to degrees celsius.<br> Need this now
yan [13]

Answer:

(97°F − 32) × 5/9 = 36.111°C

Explanation:Hope this helped

3 0
3 years ago
An electron is released from rest in a vacuum between two flat, parallel metal plates that are apart and are maintained at a con
il63 [147K]

Answer:

v = √(2qV/m)

Explanation:

½mv² = qV

mv² = 2qV

v² = 2qV/m

v = √(2qV/m)

where v = drift velocity

q = charge

V = p.d

m = mass of the electron

4 0
3 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
4 years ago
Read 2 more answers
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