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-Dominant- [34]
3 years ago
5

if spiderman runs and jumps horizontally from the top of a 200m high building, what does he need his velocity to be to land on a

building 100m high, 10m away from him?
Physics
1 answer:
antiseptic1488 [7]3 years ago
7 0

Answer:

v = 2.22 m/s

Explanation:

First we apply the second equation of motion to the vertical motion of the body:

s = Vi t + (1/2)gt²

where,

s = y = vertical distance covered = 200 m - 100 m = 100 m

Vi = V₀y = Vertical Component of Initial Velocity = 0 m/s  (because spider man jumps horizontally, thus his velocity has no vertical component initially)

t = Time Taken to Land on 100 m high building = ?

g = 9.8 m/s²

Therefore,

100 m = (0 m/s)t + (0.5)(9.8 m/s²)t²

t² = (100 m)/(4.9 m/s²)

t = √(20.4 s²)

t = 4.5 s

Now, we analyze the horizontal motion. Neglecting air friction, the horizontal motion is uniform with uniform velocity. Therefore,

s = vt

where,

s = x = horizontal distance covered = 10 m

v = V₀ₓ = Horizontal Component of Initial Velocity = Initial Velocity = ?

Therefore,

10 m = v(4.5 s)

v = 10 m/4.5 s

<u>v = 2.22 m/s</u>

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Children are sled riding on a hill One little girl pulls her sled back up the hill and does 379.5 joules of work while pulling i
Serggg [28]

Answer:

2.2N

Explanation:

Given parameters:

Work done  = 379.5J

Height  = 173m

Unknown:

Amount of force exerted on the sled  = ?

Solution:

The amount of force she exerted on the sled is the same as her weight.

Work done is the force applied to move a body through a distance.

      Work done  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

     mg  = weight;

        Work done  = weight x h

           379.5 = weight  x  173

           weight  = \frac{379.5}{173}   = 2.2N

4 0
2 years ago
Consider the electric force between a pair of charged particles a certain
Crazy boy [7]

Answer:

Doubled

Explanation:

F = (kq1q2) / r^2

F and q (Either q1 or q2) are directly proportional, so double the charge would also double the electruc force between the charges.

5 0
3 years ago
In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
Suppose that two point charges, each with a charge of +1.00 C, are separated by a distance of 1.0 m. If the distance between the
Aleksandr-060686 [28]

Given:

The magnitude of each charge is q1 = q2 = 1 C

The distance between them is r = 1 m

To find the force when distance is doubled.

Explanation:

The new distance is

\begin{gathered} r^{\prime}=\text{ 2r} \\ =2\times1 \\ =2\text{ }m \end{gathered}

The force can be calculated by the formula

F=k\frac{q1q2}{(r^{\prime})^2}

Here, k is the constant whose value is

k=9\times10^9\text{ N m}^2\text{ /C}^2

On substituting the values, the force will be

\begin{gathered} F=9\times10^9\times\frac{1\times1}{(2)^2} \\ =2.25\times10^9\text{ N} \end{gathered}

7 0
1 year ago
4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con
Inessa [10]

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km      1000 m       1 hour         1 min

----------- x ------------- x -------------- x ----------   =  44.4 m/s

1 hour            1 km         60 min      60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m          

----------   X     x sec   =  700 m

1  sec

Solve for x sec

x sec = 700m / 44.4 m/s

         =  15.8 seconds

3 0
3 years ago
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