<span>If there isn't any force then the normal contact force will be

N=m*g=7.5*9.81=73.58N

which is 73.58-23=50.58N less

so, there the person must pull at 23 degree upward

break down the tension in two components, vertical and horizontal.

vertical tension= 50.58=T*sin23

T=50.58/sin23=129.45N</span>

7 0

3 years ago

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Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o

seraphim [82]

Answer:

$\theta = 15^o \: or\: 75^o$

Explanation:

As we know that the formula of range is given as

$R = \frac{v^2sin2\theta}{g}$

now we know that

maximum value of the range of the projectile is given as

$R_{max} = \frac{v^2}{g}$

now we need to find such angles for which the range is half the maximum value

so we will have

$\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}$

$sin(2\theta) = \frac{1}{2}$

$2\theta = 30 or 150$

$\theta = 15^o \: or\: 75^o$

7 0

3 years ago

How long does it take a car to cross a 20m bridge if it starts from rest and accelerates at 5 m/s^2?

polet [3.4K]

The correct answer is 2.8s

5 0

3 years ago

You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital spee

Alchen [17]

Answer:

The minimum total speed is 11.2km/s

Explanation:

We are been asked to find the escape velocity.

Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.

The formula to calculate the escape velocity is Ve = √2gR

For the earth g = 9.8m/s2 , R = 6.4*10^6

Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s

=11.2km/s

5 0

3 years ago

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