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3 years ago

Is a glacier nothing more than a huge mineral flowing downhill?

Physics

1 answer:

NNADVOKAT [17]3 years ago

4 0

Yes. A glacier is nothing more than a huge mineral flowing downhill.

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What force must be applied to the end of a rod along the x-axis of length 2.25 m in order to produce a torque on the rod about t

4vir4ik [10]

Answer:

<h2>4.9N approx.</h2>

Explanation:

Step one:

given data

length of rod r= 2.25m

Required torque = 11 Nm

Required

The force needed to produce a torque of 11Nm

Step two:

<em>"By definition, Torque is the twisting force that tends to cause rotation. </em>

<em>The point where the object rotates is known as the axis of rotation."</em>

Mathematically,

$\tau = rF\sin\theta$

$\tau$ = torque

r = radius

F = force

$\theta$ = angle between F and the lever arm

in this case is zero

$\tau = rF\\\\F=\tau/r$

substituting we have

F=11/2.25

F=4.88N

F=4.9N approx.

7 0

2 years ago

The image below The image below The image below The image below The image below The image below The image below The image below

larisa [96]

Answer:

wow i have know idea

Explanation:

5 0

2 years ago

1. Is the image projected on a movie screen real or virtual? What about the image of yourself seen in a bathroom mirror?

antiseptic1488 [7]

Answer: 1. The movie one is virtual and the bathroom mirror is real

2. The image is distorted in a way

3. Behind the spoon

Explanation:

7 0

1 year ago

Two forces of magnitudes 4.0 Newtons and 3.0 Newtons pull on a box. The forces make an angle of 40° with each other. What is the

leonid [27]

Answer:

6.6 N

Explanation:

Let's take the direction of the force of 4.0 N as positive x-direction. This means that the force of 3.0 N is at 40 degrees above it. So the components of the two forces along the x- and y-directions are:

$F_{1x} = 4.0 N\\F_{1y} = 0$

$F_{2x} = 3.0 N cos 40^{\circ}=2.3 N\\F_{2y} = 3.0 N sin 40^{\circ} = 1.9 N$

So the resultant has components

$F_x = F_{1x}+F_{2x}=4.0 N +2.3 N = 6.3 N\\F_y = F_{1y} + F_{2y} = 0 + 1.9 N = 1.9 N$

So the magnitude of the resultant is

$F=\sqrt{F_x^2 +F_y^2}=\sqrt{(6.3)^2+(1.9)^2}=6.6 N$

And in order for the body to be balanced, the third force must be equal and opposite (in direction) to this force: so, the magnitude of the third force must be 6.6 N.

3 0

2 years ago

At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1400 kg car traveling at 6

Ilia_Sergeevich [38]

Momentum of car

Given: Mass m= 1,400 Kg; V = 6.0 m/s

Formula: P = mv

P = (1,400 Kg)(6.0 m/s)

P = 8,400 Kg.m/s

Velocity of the rider to have the same momentum as a car.

Mass of rider and bicycle m = 100 Kg

P = mv

V = P/m

V = 8,400 Kg.m/s/100 Kg

V = 84 m/s

7 0

2 years ago