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jasenka [17]
2 years ago
11

There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2

and θ= 30◦, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis.
Physics
1 answer:
maw [93]2 years ago
8 0

Answer:

second force = 32.784

Magnitude =\sqrt{32.784

θ = -90°

Explanation:

a)

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

    = 24 × ( 1 + √3 )÷ 2

  =12 (1 +√3 )

  = 32.784

b) \sqrt{12(1 +\sqrt{3}}

= \sqrt{12 ( 1+ 1.732)}

= \sqrt{12 (2.732)}

= \sqrt{32.784}  

c)

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

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kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, B=Vg\rho where V is volume, g is gravitational constant and \rho is density

B+F_{upward}=mg where F_{upward} is upward force

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F_{upward}=mg- Vg\rho_{w}

F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}} where \rho_{hippo} is the density of hippo

F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})

Using g as 9.81

F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N

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3 0
2 years ago
Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.
sineoko [7]

Answer:

Total number of electrons

N = 7.25 \times 10^{24}

electrons removed from each sphere

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = 7.27 \times 10^{-10}

Explanation:

As we know that moles is defined as

n =\frac{mass}{molar mass}

n = \frac{0.0250}{0.026982}

n = 0.93

so number of atoms of Al in each sphere is given as

N = 0.93(6.02 \times 10^{23})

N = 5.58 \times 10^{23}

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

N = 13 \times (5.58 \times 10^{23})

N = 7.25 \times 10^{24}

Also we know that force of attraction between them is given as

F= \frac{kq_1q_2}{r^2}

1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}

q = 8.4 \times 10^{-4} C

now we have

q = Ne

8.4 \times 10^{-4} = N(1.6 \times 10^{-19}

N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}

N = 5.27 \times 10^{15}

Fraction of electrons transferred is given as

f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}

f = 7.27 \times 10^{-10}

6 0
2 years ago
A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de
docker41 [41]

Answer:

Magnetic field at the center of the loop B=5.89\times 10^{-5}\ T.

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.

We know , magnetic field at the center of loop is given by :

B=N\dfrac{\mu_o i}{2r}

Putting all values in above equation we get :

B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.

Hence , this is the required solution.

8 0
2 years ago
Determine the sign (+ or −) of the torque about the elbow caused by the biceps, τbiceps, the sign of the weight of the forearm,
Alex Ar [27]
Ans: 
1.  τbiceps = +(Positive)
2.  τforearm = -(Negative)
3.  τball = -(Negative)

Explanation:

The figure is attached down below.

1. T<span>orque about the elbow caused by the biceps, τbiceps:
Since Torque = r x F (where r and F are the vectors)
</span>Where r is the vector from elbow to the biceps.
<span>
We can see in the figure that F(biceps) is in upward direction, and by applying the right hand rule from r to F, we get the counterclockwise direction. The torque in counterclockwise direction is positive(+). Therefore, the sign would be +.

2. </span>Torque about the the weight of the forearm, τforearm:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the forearm.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(forearm) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

3. Torque about the the weight of the ball, τball:
Since Torque = r x F (where r and F are the vectors)
Where r is the vector from elbow to the ball.

Also weight is the special kind of Force caused by the gravity.

We can see in the figure that W(ball) is in downward direction, and by applying the right hand rule from r to F, we get the clockwise direction. The torque in clockwise direction is negative(-). Therefore, the sign would be -.

8 0
2 years ago
A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second
Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
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