How does the construction of dams positively affect natural resources?

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0

4 years ago

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3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

nikklg [1K]

Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)

and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

$a = \frac{80 m/s - 0 m/s}{0.05 sec} = 1600 \frac{m}{s^2}$

Notice that the units of acceleration in the SI system are $\frac{m}{s^2}$ (meters divided square seconds)

7 0

3 years ago

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A cricket player catches the ball leaning towards to the ground,why?

Vadim26 [7]

Answer:

Explanation:

As it’s difficult to catch it from up.

Gravitational force will pull us when we jump.

If gravity was not there, he could catch the ball. But he will float in the sky after that.

That’s the answer

3 0

4 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

= $5 \times 10^{-5}$

3 0

4 years ago

What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o

leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,

$d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}$

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

$\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}$

This value can be used to determine the time for the pellet through the barrel:

$t = \frac{d}{\overline{v}}=\frac{0.6m}{77.5\frac{m}{s}}\approx0.00774s$

Finally, we can use the above to calculate the force:

$F = ma = m\frac{2d}{t^2} = 0.007kg\cdot \frac{2\cdot 0.6 m}{0.00774^2 s^2}\approx 140.22N$

8 0

3 years ago