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myrzilka [38]
1 year ago
7

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of

the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70-kg pilot at that point?
Physics
1 answer:
adoni [48]1 year ago
6 0

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

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"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression
zavuch27 [327]

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

6 0
2 years ago
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with
Lana71 [14]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        Em_{f} = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     Em_{f} = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

7 0
2 years ago
How much is the moment due to force P about point B. P(unit=N) vector is equal to (150i+260j ) and vector BA (unit=meter) is equ
artcher [175]

Answer:

The moment (torque) is given by the following equation:

\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)

Explanation:

The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.

4 0
1 year ago
A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm
Alisiya [41]

Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{12}ml^2+mx^2

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2

I=0.7500\ kg-m^2

Hence, The moment of inertia is 0.7500 kg-m².

5 0
1 year ago
Marcus can drive his boat 24 miles down the river in 2 hours but takes 3 hours to return upstream. Find the rate of the boat in
notsponge [240]

Answer:

speed of boat as

v_b = 10 mph

river speed is given as

v_r = 2 mph

Explanation:

When boat is moving down stream then in that case net resultant speed of the boat is given as

since the boat and river is in same direction so we will have

v_1 = v_r + v_b

Now when boat moves upstream then in that case the net speed of the boat is opposite to the speed of the river

so here we have

v_2 = v_b - v_r

as we know when boat is in downstream then in that case it covers 24 miles in 2 hours

v_1 = \frac{24}{2} = 12 mph

also when it moves in upstream then it covers same distance in 3 hours of time

v_2 = \frac{24}{3} = 8 mph

v_b + v_r = 12 mph

v_b - v_r = 8 mph

so we have speed of boat as

v_b = 10 mph

river speed is given as

v_r = 2 mph

8 0
1 year ago
Read 2 more answers
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