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3 years ago

12

john doe operates a crane that can pick up 3.0 tons of excavated earth in an hour. John's wages are $35 per hour. What, then, is

the cost of picking up 85kg of excavated earth?

1 answer:

levacccp [35]3 years ago

8 0

Answer:

1

Explanation:

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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10

Charra [1.4K]

Answer:

$u(t)=1.15 \sin (8.68t)cm$

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s$

Where $g=980 cm/s^2$

$u(t)=Acos8.68 t+Bsin 8.68t$

u(0)=0

Substitute the value

$A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t$

Substitute u'(0)=10

$8.68B=10$

$B=\frac{10}{8.68}=1.15$

Substitute the values

$u(t)=1.15 \sin (8.68t)cm$

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0

3 years ago

2. What is the feather's initial velocity (before it is dropped) in m/s?

Dmitrij [34]

Answer: 0

Explanation: Initial velocity is 0.

7 0

3 years ago

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

3 years ago

8. A driver starts his parked car and within 12 hours reaches a speed of 75km/h, as he travels

erastovalidia [21]

Answer:

56mph

Explanation:

7 0

2 years ago

When you drive through deep water, you should dry out your brakes by _____________. pumping your brakes and driving in low gear

Mandarinka [93]

Answer:

By Applying pressure to the brakes

Explanation:

Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.

If the brakes are wet test them by pressing or tapping on them gently.

You can as well dry brakes by driving in low gear and applying pressure to the brakes.

7 0

3 years ago