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Viktor [21]
1 year ago
7

A 6-ft-long fishing rod AB is securely anchored in the sand of a beach. After a fish takes the bait, the resulting force in the

line is 8.6 lb. Determine the moment about A of the force exerted by the line at B. (Round the final answer to one decimal place.)
Physics
1 answer:
babunello [35]1 year ago
4 0

Answer:

Moment of Force, M = 51.6 lbf-ft

Explanation:

Given,

The fishing rod AB of length, L = 6 ft

The force acting on the rod, F = 8.6 lbf

The moment of force about A is given as the product of force and perpendicular distance. It is given by the formula

                                    M = F x L  

Substituting the given values in the above equation

                                     M = 8.6 lbf  x 6 ft

                                         = 51.6 lbf-ft

The moment about A of the force exerted by the line at B is, M = 51.6 lbf-ft

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet
Gnom [1K]

Answer:

d. 37 °C

Explanation:

m_{m} = mass of lump of metal = 250 g

c_{m} = specific heat of lump of metal  = 0.25 cal/g°C

T_{mi} = Initial temperature of lump of metal = 70 °C

m_{w} = mass of water = 75 g

c_{w} = specific heat of water = 1 cal/g°C

T_{wi} = Initial temperature of water = 20 °C

m_{c} = mass of calorimeter  = 500 g

c_{c} = specific heat of calorimeter = 0.10 cal/g°C

T_{ci} = Initial temperature of calorimeter = 20 °C

T_{f} = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) +  m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C

6 0
2 years ago
The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo
agasfer [191]

As we know by Doppler's effect of light we have

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

3.01 \times 10^{-3} = \frac{v}{c}

here we have

v = 904.3 km/s

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

3 0
1 year ago
Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an
Svet_ta [14]
Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
(velocity)^2 = 10.78
velocity = 3.28 meters/sec
8 0
2 years ago
Calculate the average charge on arginine when ph=9.20. (hint : find the average charge for each ionizable group and sum these to
weqwewe [10]
Arginine is a basic aminoacid, because it has two amino groups and one acid group. At a low pH, every ionizable group is protoned. At a little higher pH, the acid group looses its proton. A little higher pH, one amino group looses its proton. At a very high pH, all ionizable groups are not protoned.

Pkas

<span> <span><span> <span> pka1 = 1.82 </span> <span> pka2 = 8.99 </span> <span> pka3 = 12.48 </span> </span> </span></span> So 9.20 is higher tan the second pKa and lower than the third pka. This means the acid has already lost its proton, and one of the aminos too, but the second amino hasn’t. When an acid is not protoned, it has a negative charge. When an amino is not protoned, it’s neutral. When an amino is protoned, it has a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative charge (the acid), what makes it neutral.
4 0
1 year ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
1 year ago
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