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zhannawk [14.2K]
3 years ago
4

Two reindeer-in-training pull on a sleigh. Connie pulls with a force of 200 N at an angle of 20 degrees above the (positive) -ax

is, while Randolph pulls with a force of 800 N at an angle of 50 degrees ° below the positive) -axis. What is their resultant magnitude of the force on the sleigh?
Physics
1 answer:
salantis [7]3 years ago
7 0

Answer:

F_r \approx 978.3N

\theta =44 \textdegree

Explanation:

From the question we are told that

Connie pulls with a force of F_c=200 N

At an angle \theta _c=20 \textdegree

Randolph pulls with a force of F_R=800 N

At an angle \theta _R=50 \textdegree

Generally the horizontal axis  can mathematically be represented as

f_x=f_ccos\theta _c+F_Rcos\theta _R

f_x=200*cos20+800*cos50

f_x=702.2N

Generally the vertical axis  can mathematically be represented as

f_y=f_csin\theta _c+F_Rsin\theta _R

f_y=200*sin20+800*sin50

f_y=681.24N

Generally the resultant force F_r is mathematically  given by

F_r=\sqrt{f_x^2+f_y^2}

F_r=\sqrt{702.1686119^2+681.2395832^2}

F_r=978.3230731N

F_r \approx 978.3N

Generally the Direction \theta of force is mathematically given by

\theta =tan^-^1(\frac{f_c}{f_y})

\theta =tan^-^1(\frac{681.23}{702.16})

\theta =44 \textdegree

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aivan3 [116]

Answer:

The fringes are 4.7*10^-7 m apart, such that they are adjacent.

Explanation:

Using the formula for adjacent fringes given a single slit:

Δx=\frac{(Wavelength)(Distance between slit and screen)}{Width}

Δx=\frac{(590/10^{9})(1.90) }{(2.30)}

Δx=0.000000487 m

Hope this helps!

7 0
2 years ago
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A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a
anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = 4.83 \AA

Intercepts along the three axes:

\bar{x} = 9.66 \AA

\bar{x} = 19.32 \AA

\bar{x} = 14.49 \AA

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = \frac{1}{9.66}

b = \frac{1}{19.32}

c = \frac{1}{14.49}

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = \frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}

b' = \frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}

c' = \frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}

7 0
3 years ago
A 50.0 kg object is moving at 18.2 m/s when a 200 N force is applied opposite the direction of the objects motion, causing it to
Gekata [30.6K]

Answer:

t = 1.4[s]

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

P=m*v\\or\\P=F*t

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 50 [kg]

v = velocity [m/s]

F = force = 200[N]

t = time = [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

(m_{1}*v_{1})-F*t=(m_{1}*v_{2})

where:

m₁ = mass of the object = 50 [kg]

v₁ = velocity of the object before the impulse = 18.2 [m/s]

v₂ = velocity of the object after the impulse = 12.6 [m/s]

(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]

3 0
3 years ago
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through
Svet_ta [14]

Answer:

The speed is \sqrt{2}v_{0}.

(a) is correct option.

Explanation:

Given that,

Potential difference V= V_{0}

Speed v = v_{o}

If it were accelerated instead

Potential difference V'=2V_{0}

We need to calculate the speed

Using formula of initial work done on proton

W = q V

We know that,

\Delta W=\Delta K.E

q V=\dfrac{1}{2}mv^2

Put the value into the formula

q V_{0}=\dfrac{1}{2}mv_{0}^2

v_{0}^2=\dfrac{2qV_{0}}{m}....(I)

If it were accelerated instead through a potential difference of 2 V_{0}, then it would gain a speed will be given as :

Using an above formula,

v_{0}'^2=\dfrac{2qV_{0}}{m}

Put the value of V_{0}

v_{0}'^2=\dfrac{2q\times2V_{0}}{m}

v_{0}'=\sqrt{\dfrac{4qV_{0}}{m}}

v_{0}'=\sqrt{2}v_{0}

Hence, The speed is \sqrt{2}v_{0}.

6 0
3 years ago
I tried to solve for it 3 times but I still got it wrong. HELP?!?!
QveST [7]
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