A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?

A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi

kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well nitrate per 15 $\mathrm{m}^{3}$ of water

nitrate in distant reservoir $=5 \mathrm{~g} / \mathrm{m}^{3}$

Let the flow rate of well

flow rate of reservoir $=y m^{3} / s$

Drinking water requirement is $45 \mathrm{ppm}$ or $45 \mathrm{~g} / \mathrm{m}^{3}$

therefore, the total flow of drinking water

8 0

2 years ago

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

Lana71 [14]

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the given Normal on block 2 = N

and $N = m_2 g \cos \theta$

and the tension in the given string is said to be $T = m_2 g \sin \theta$

When the acceleration $a=\frac{F}{m_1}$

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

$F = m_1g-T.$

It is known that if the said

$\theta=\frac{\pi}{2}$ ,

then Tension $T= m_2g$ $[since \sin(\pi/2) = 1]$ ,

Now making $"F = m_1g - m_2g"$

So If we are to make Force equal to zero

$F=0 => m_1g = m_2g \ or \ m_1 = m_2$

6 0

2 years ago

Assume we’re able to travel to your planet and decide to take some fireworks with us to celebrate our journey.

julia-pushkina [17]

Answer:

The horizontal distance covered by the firework will be $\frac{1876.8}{g}$

Explanation:

Let acceleration due to gravity on the planet be g, initial velocity of the firework be u and angle made with the horizontal be ∅.

writing equation of motion in vertical direction:

$v_{y}=u_{y}+(-g) t$

$u_{y}= u\sin \phi$

and $v_{y}=0$

therefore $\frac{u\sin \phi }{g} =t$

writing equation of motion in horizontal direction:

$s_{x}=u_{x}t$

$u_{x} = u\cos \phi$

therefore the equation becomes $s_{x}=\frac{u^{2} \sin \phi \cos \phi}{g}$

therefore horizontal distance traveled = $\frac{u^{2}\sin 2\alpha \phi }{2g}=\frac{1876.8}{g}\frac{m}{s}$

5 0

3 years ago

1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the

zheka24 [161]

Answer:

a) F = 4.9 10⁴ N, b) F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

∑F = 0

F-W = 0

F = W = mg

F = 5000 9.8

F = 4.9 10⁴ N

the area of the pisto is

A = pi r²

A = pi d² / 4

A = pi 1 ^ 2/4

A = 0.7854 m²

pressure is

P = 4.9 104 / 0.7854

P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

where subscript 1 is for the small piston and subscript 2 is for the large piston

F₁ = $\frac{A_1}{A_2} \ F_2$

the force applied must be equal to the weight of the truck

F₁ = $( \frac{d_1}{d_2} )^2\ m g$

F₁ = (0.05 / 1) ² 5000 9.8

F₁ = 122.5 N

7 0

2 years ago

If two people are running at the same speed in the same direction. One person is one meter ahead of the other. The person in fro

Gre4nikov [31]

Answer:yes

Explanation:

5 0

2 years ago