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Kamila [148]
3 years ago
6

What is the maximum weight a boat can hold if a boat can displace 60.5ml?

Physics
1 answer:
Valentin [98]3 years ago
6 0

Answer:

a. 60.5 kg

Explanation:

Given data,

The maximum water a boat can displace is, 60.5 ml

According to the principle of buoyancy, the weight of the floating body is equal to the weight of the liquid displaced.

Under standard temperature and pressure, a unit mass of water equals one liter.

If a boat can displace a maximum of 60.5 ml of water, then it can hold a mass of a maximum of 60.5 kg of mass.

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You have a 1.55 kg block of silver at room temperature of 20 °C. Silver melts at 962°C. How much heat energy would be needed to
notsponge [240]
Specific heat capacity= heat energy/mass×temperature rise

962°C - 20°C = 942K

Heat energy (Eh) = 239 × 1.55 × 942
Eh= 348963.9J

shc of Ag: 238.6 J/kg-K

m of Ag: 1.55kg
5 0
3 years ago
A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force<br> acting on it.
AleksandrR [38]
Force = mass*acceleration so
3.6*2.5 =9 Newtons
5 0
3 years ago
Suppose a photon with an energy of 1.60 eV strikes a piece of metal. If the electron that it hits loses 0.800 eV leaving the met
JulsSmile [24]

To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

\Delta E = E_0 -W_f

Where

E_0 =Initial Energy

W_f = Energy loses

Replacing we have that

\Delta E = 1.6-0.8

\Delta E = 0.8eV

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

7 0
3 years ago
Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
lapo4ka [179]

Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

Spring constant of each spring = k

Thus,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

7 0
3 years ago
Read 2 more answers
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m
slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 m/s^{2} and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

M\frac{dv}{dt} = u\frac{dM}{dt}

\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt

       = u\int_{M_{o}}^{M_{f}} \frac{dM}{M}

v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}

v_{o} = 0

As, v_{f} = u ln (\frac{M_{f}}{M_{o}})

u = -2900 m/s

M_{f} = M_{o} - m \times t_{f}

           = 2500 kg + 1000 kg - 95 kg \times t_{f}s

           = (3500 - 95t_{f})s

v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s

Also, we know that

     a = \frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}

        = \frac{u}{3500 - 95 t} \times (-95) m/s^{2}

        = \frac{95 \times 2900}{3500 - 95t} m/s^{2}

At t = 10 sec,

v_{f} = 918.34 m/s

and,   a = 108 m/s^{2}

3 0
3 years ago
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