All categories

3 years ago

The amount of force needed to keep a 0.1 kg basketball moving at a constant speed of 6 m/s on frictionless ice is how much N?

Physics

1 answer:

FromTheMoon [43]3 years ago

3 0

Mass=m=0.1kg
Acceleration=0(Constant speed)

$\\ \sf\longmapsto F=ma$

$\\ \sf\longmapsto F=0.1(0)$

$\\ \sf\longmapsto F=0N$

You might be interested in

When ultra violets lights shine on glass what does it do to electrons in the glass structure?

andreev551 [17]

Answer:

Explanation:

7 0

3 years ago

Cells that remain as sources of other cells are called

inna [77]

They are called stem cells. This cells are undifferentiated which means it can specialize in other types when it receives the right stimuli. They can divide through mitoses and become more stem cell or become a bone, muscle, blood cell, etc.

They can have 2 origins: embryos or some human tissue; their function is to regenerate or substitute damaged cells

8 0

3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou

Vilka [71]

The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2

3 0

4 years ago

A person throws a balloon straight down above level ground. The balloon bounces back up into the air. Drag force is significant

love history [14]

Explanation:

a) Balloon is being thrown down and is speeding up;

mg > $F_drag$

b) Balloon is in the air on its way down and moving with constant speed.

$F_drag$ =mg

c) The Balloon is on the ground and rest instantaneously

mg = Normal

d) Balloon is moving slowly downward;

e) Because, at the peak of trajectory drag force is 0.

Drag force = 0

5 0

3 years ago