A kangaroo jumps up with an initial velocity of

A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti

aleksley [76]

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ = 30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

7 0

2 years ago

A football player kicks the ball at a 45 degree angle. Without an effect from the wind, the ball would travel 60.0m horizontally

Ronch [10]

B when the ball is at its maxium height

5 0

2 years ago

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

5 0

2 years ago

A person who weighs 800N on the earth's surface will weigh 200N at what height above the earth

Marina86 [1]

Answer: 6,400 km

Explanation:

The weight of a person is given by:

$W=mg$

where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:

$g=\frac{GM}{r^2}$

where

G is the gravitational constant

M is the Earth's mass

r is the distance of the person from the Earth's center

The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):

$800 N= W=mg=m \frac{GM}{R^2}$ (1)

Now we want to find the height h above the surface at which the weight of the man is 200 N:

$200 N = W' = mg' = m \frac{GM}{(R+h)^2}$ (2)

If we divide eq.(1) by eq.(2), we get

$\frac{800 N}{200 N}=\frac{W}{W'}=\frac{(R+h)^2}{R^2}$

$4=\frac{(R+h)^2}{R^2}$

By solving the equation, we find:

$4R^2 = (R+h)^2=R^2+2Rh+h^2\\h^2 +2Rh-3R^2 =0$

which has two solutions:

$h=-3R$ --> negative solution, we can ignore it

$h=R$ --> this is our solution

Since the Earth's radius is $R=6.4\cdot 10^6 m$ , the person should be at $h=R=6.4\cdot 10^6 m=6400 km$ above Earth's surface.

5 0

3 years ago

Two piles of heavy boxes must be lifted onto a shelf. Each pile has 20 boxes and each box is the same size and weight. Tom and c

Murrr4er [49]

Tom used more Force but over a shorter distance. Tom and Claudia both did the same amount of work.

4 0

3 years ago

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