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lidiya [134]
3 years ago
13

How does the average number of drops required to break open a whelk depend on platform height?

Physics
1 answer:
Hunter-Best [27]3 years ago
8 0
The correct answer for this question is this one: "The number of drops required decreases."<span>The average number of drops required to break open a whelk depend on platform height decreases as the experiment progresses.</span>
Hope this helps answer your question and have a nice day ahead.
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Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 194 m, at what angle should th
Brums [2.3K]

Answer:

The banking angle is 23.84 degrees.

Explanation:

Given that,

Radius of the curve, r = 194 m

Speed of the car, v = 29 m/s

On the banked curve, the centripetal force is balanced by the force of friction such that,

mg\ tan\theta=\dfrac{mv^2}{r}

tan\theta=\dfrac{v^2}{rg}

tan\theta=\dfrac{(29)^2}{194\times 9.8}

\theta=23.84^{\circ}

So, the banking angle is 23.84 degrees. Hence, this is the required solution.    

5 0
3 years ago
SO MANY POINTS! WILL MARK BRAINIEST!!!! PLZ BE FASSTT!
lora16 [44]

Answer:  Point A is the answer for the potential energy. Point D is the answer for the kinetic energy.

Explanation:

4 0
3 years ago
A 2000 kg truck is traveling at 5 m/s and collides with a 1000 kg car that is not moving. After the collision, the 2000 truck st
sp2606 [1]

Answer:

A) 10 m/s

Explanation:

We know that according to conservation of momentum,

m1v1 + m2v2 = m1u1 + m2u2  ..............(equation 1)

where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.

From the given data

If truck and car are two bodies

truck :       m1 = 2000 Kg           v1 = 5 m/s                u1 = 0

car    :        m2 = 1000 kg           v2 = 0                      u2 = ?

final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.

substituting the values in equation 1, we get

(2000 x 5) + 0 = 0 + (1000 x u2)

u2 = \frac{2000}{1000} x 5

    = 10 m/s

Hence after collision, car moves at a velocity of 10 m/s

3 0
3 years ago
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
3 years ago
A package was determined to have a mass of 5.7 kilograms. What's the force of gravity acting on the package on earth?
Arturiano [62]
1.) The force of gravity is what we call weight, we define it as:
w=mg
w=5,7kg*9,8m/s²
w=55,86kg (b)
2.) We know that:
power=W/t
power=50J/20s
power=2,5Watts (a)
3.) The work done is equal to the potential energy, so:
Epg=mgh
Epg=63kg*9,8m/s²*7m
Epg=4321J
Now we get the power:
power=W/t
power=4321J/5s
power=864Watts
Now:
1HP=746Watts
=1,16HP (b)
4.) We know that:
F=ma
350N=m*10m/s²
m=350N/10m/s²
m=35kg (b)
5.) d.) Aceleration is tha rate of change in velocity, either positive (increasing) or negative (decreasing)
8 0
3 years ago
Read 2 more answers
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