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Nutka1998 [239]
3 years ago
9

A small ferryboat is 3.90 m wide and 6.30 m long. when a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t

he river. what is the weight of the truck?
Physics
1 answer:
Leya [2.2K]3 years ago
6 0
When the truck's weight is added to the boat, the boat sinks 5 cm deeper,
and displaces additional water whose weight is equal to the weight of the
truck.

The volume of the additional displaced water is

             (3.9 m) x (6.3 m) x (5.0 cm)

         =  (3.9 m) x (6.3 m) x (0.05 m)  =   1.2285 m³ .

The weight of that much water is the weight of the truck.

          Mass of 1 liter of water  =  1 kilogram

          1.2285 m³  =  1,228.5 liters  =  1,228.5 kg of water.

          Weight = (mass) x (gravity)

                      = (1,228.5 kg) x (9.8 m/s²)  =  12,039 Newtons.

                                                                  (about 2,708 pounds) 

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A e limb of mass 12kg falls straight down. If air resistance exerts 27 N of froce on the limb as it falls, what is the net force
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7 0
3 years ago
Read 2 more answers
A non-ideal 12.2 V battery is connected across a resistor R. The internal resistance of the battery is 1.9Ohm. Calculate the pot
Brums [2.3K]

Answer:

R=100 Ohm, V=11.97 volts and I=0.12 amperes

R=10 Ohm, V=10.25 volts and I=1.20 amperes

R=2 Ohm, V=6.26 volts

Explanation:

The potential difference (voltage) of a battery with internal resistance is:

V=\xi-Ir (1)

with \xi the electromotive force (the voltage the batteries say to has) , I the current and r the internal resistance. By Ohm's law the current that passes through the resistor is:

I=\frac{V}{R} (2)

using (2) on (1):

V=\xi-\frac{V*r}{R}

solving for V:

V+\frac{V*r}{R}=\xi

V=\frac{\xi}{1+\frac{r}{R}} (3)

R=100 Ohm

V=\frac{12.2}{1+\frac{1.9}{100}}=11.97 V

R=10 Ohm

V=\frac{12.2}{1+\frac{1.9}{10}}=10.25 V

R=2 Ohm

V=\frac{12.2}{1+\frac{1.9}{2}}=6.26 V

Because we have now the values of I on the circuit (is the same through all the components because is a series circuit)

We use back substitution on (1) to find the current:

R=100 Ohm

I=\frac{V}{R}=\frac{11.97}{100}=0.12 A

R=10 Ohm

I=\frac{V}{R}=\frac{11.97}{10}=1.20 A

7 0
3 years ago
A pitcher throws a baseball with a mass of 143 g horizontally at a speed of 38.8 m/s (87 mi/h). The hitter's bat is in contact w
SIZIF [17.4K]

Answer:

F = −10093.41 N

Explanation:

Given that,

Mass of a baseball, m = 143 g = 0.143 kg

Initial speed of the baseball, u = +38.8 m/s

The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s

We need to find the average force exerted on the ball by the bat. So, Force is given by :

F=ma

a is acceleration

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.143\times (-45.9-(38.8))}{1.2\times 10^{-3}}\\\\F=-10093.41\ N

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.

8 0
3 years ago
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