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Stolb23 [73]
3 years ago
12

Josh is refilling his pool with clean water. How many gallons of water will it take if the pool is 50m by 25m by 1.5m?

Physics
1 answer:
horsena [70]3 years ago
7 0

Answer:

V = 493421.05 [gal]

Explanation:

This is a problem that consists of handling units, we can calculate by first-hand the volume, then convert units from cubic meters to gallons.

V = 50 * 25 * 1.5

V = 1875 [m^3]

Now we need to convert units, using the proper conversion factor.

1875[m^3]*\frac{1000lt}{1m^3} *\frac{1gal}{3.8lt} \\493421.05[gal]

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3 years ago
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parall
dolphi86 [110]

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

W_P=Px

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

W_P=(171N)(8.80m)=1504.8J

(b) The work don by the friction force is:

W_f=F_fx=\mu N x=\mu Mg x

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J

(c) The Normal force is

N=Mg=(46.0kg)(9,8m/s^2)=450.8N

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

W_N=0J

(d) the same as before:

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8 0
3 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

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3 years ago
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Answer:

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