F = M x G<br><br> Find the force of gravity acting upon a 1500Kg Hippopatamus.

Three ropes A, B and C are tied together in one single knot K. (See figure.)

noname [10]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

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3 0

2 years ago

g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?

kiruha [24]

Answer:

Range will become 4 times of initial range

Explanation:

Let the velocity of projection is u

And angle at which projectile is projected is $\Theta$

And acceleration due to gravity is $g\ m/sec^2$

So range of projectile is equal to $R=\frac{u^2sin2\Theta }{g}$ ........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity $u_{new}=2u$

So new range will be equal to $R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}$ .....eqn 2

Now dividing eqn 2 by eqn 1

$\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }$

$R_{new}=4R$

So if we double the initial launch speed then range will become 4 times

4 0

4 years ago

A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.

kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

$\displaystyle d=v.t$

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

$\displaystyle t=\sqrt{\frac{2*20}{9.8}}$

$\displaystyle t=\sqrt{4.0816}$

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

$\displaystyle d=10\cdot 2.02$

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago

Calculate kinetic energy of a planet using 5.97 x 10^24 kg mass and v at 30.29 km s-1

kiruha [24]

Answer:

2.74 × 10^33 J

Explanation:

the formula to calculate kinetic energy is:

1/2mv²

m= mass (kg)

v= velocity (m/s)

given that,

m = 5.97 × 10^24

v = 30.29 km s-1

= 30290 m s-1

1/2× 5.97 × 10^24 × 30290²

=2.74 × 10^33 J

4 0

2 years ago

F = M x G Find the force of gravity acting upon a 1500Kg Hippopatamus.