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Pani-rosa [81]
3 years ago
6

3. A 40.0-kg wagon is towed up a hill inclined at 18.5 with respect to the horizontal. The tow rope is parallel to the incline a

nd has a tension of 140 N. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 80.0 m up the hill?
Physics
1 answer:
inn [45]3 years ago
4 0
Force=tension-fg sin ∅
=140-mg sin 18.5
=140-124.35
=15.62N

a=f/m=15.62/40=0.39
now,
v²=u²+2as
=2×0.39×80
v²=62.4
v=7.8m/s
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A solar cooker, really a concave mirror pointed at the sun, focuses the sun's rays 16.0 cm in front of the mirror. what is the r
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The focal point of a concave mirror is halfway along the radius, therefore the radius would be 2•16= 32 cm
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3 years ago
A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 – and lands at bottom of the
Marrrta [24]

Question:

A skateboarder jumps horizontally off the top of a staircase at a speed of 14.5 and lands at bottom of the stairs. The staircase has a horizontal length of 8.00 m. We can ignore air resistance. What is the skater's vertical displacement during the jump?

Answer:

y = 1.48 m

Explanation:

Projectile motion is a two dimensional motion experienced by an object or particle that is subjected near the Earth's surface and moves along a curved path under the influence of gravity only. The path followed by projectile motion is called projectile path.

As the skateboarder followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

Also Skateboard jumps horizontally, So initial velocity has only horizontal component.

Horizontal component of initial velocity = v_{i_{x}} = 14.5 m/s

Horizontal displacement = x = 8.00 m

Vertical displacement = y = ?

Using the following formula

x = v_{i_{x}}t

8 = (14.5)(t)

t = 0.55 s

As skateboarder jumps horizontally, So there is no vertical component of velocity.

According to 2nd equation of motion

y = v_{i_{y}}t + \frac{1}{2}gt^{2}

As v_{i_{y}} = 0

So

y = 0.5gt²

y = 0.5*9.8*(0.55)²

y = 1.48 m

8 0
3 years ago
Read 2 more answers
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 9.1 kN. Determine the requ
Oksana_A [137]

Answer:

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Explanation:

Given:

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Find:

- Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A

- Determine the magnitude R of this downward force.

Solution:

- Compute the three angles as shown in figure attached, a, B , y:

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                                 B = arctan (50/30) = 59.04 degrees

                                 y = 180 - 38.36 = 82.6 degrees

- Use cosine rule to calculate R and F_ac as follows:

                                 sin(a) / T_ac = sin(B) / T_ab = sin(y) / R

                                 sin(38.36) / T_ac = sin(59.04) / 9.1 = sin(82.06) / R

                                 T_ac = 9.1 * ( sin(38.36) / sin(59.04) )

                                T_ac = 6.586 KN

                                 R = 9.1 * ( sin(82.06) / sin(59.04) )

                                 R = 10.51 KN

7 0
3 years ago
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AnnZ [28]

Answer:

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5 0
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timurjin [86]

Answer:

q = 0.384 C

Explanation:

The total charge present at the head can be easily found out by multiplying the charge on a single electron with the total number of electrons present on the head:

q = ne

where,

q = total charge on head = ?

n = total no. of electrons on the head = 2.4 x 10¹⁸

e = charge on 1 electron = 1.6 x 10⁻¹⁹ C

Therefore,

q = (2.4\ x\ 10^{18})(1.6\ x\ 10^{-19}\ C)

<u>q = 0.384 C</u>

8 0
3 years ago
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