All categories

2 years ago

VERY URGENT GIVING BRAINLY

Physics

1 answer:

lakkis [162]2 years ago

8 0

Answer:

Explanation:

There is a mechanical advantage in this system....

she only needed 100 N to lift 500 N

1:5 she lifted it 2 meters, so if there WAS NO friction, she would have to pull in 5 x 2 = 10 meters

but there IS friction so she must have pulled in MORE than 10 m

You might be interested in

Which of the following is the center of the universe according to the geocentric model?

Dima020 [189]

Earth is the center according to the geocentric model.

4 0

4 years ago

Read 2 more answers

katen-ka-za [31]

Answer: B = 0.1 Tesla

Explanation: Given that the

Emf = 6.0 V

Length L = 30 m

Speed V = 2 m/s

B = ?

The induced EMF = BVL

6 = 30 × 2 × B

B = 6/60

B = 0.1 Tesla

The magnetic field B = 0.1 Tesla.

8 0

4 years ago

Karen drops a 2.0 kg rock from the top of a mountain. Determine the distance the rock travels after falling for 3 seconds assumi

Sophie [7]

The motion of the rock is a free-fall motion, which is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ downward, so the distance covered by the rock in a time t can be found by using the formula

$S=\frac{1}{2} gt^2$

By substituting t=3 s into the formula, we find the distance the rock travels after falling for 3 seconds:

$S=\frac{1}{2} (9.8 m/s^2)(3 s)^2=44.1 m$

7 0

3 years ago

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

4 years ago

The drag is proportional to the square of the speed of the boat, in the form Fd=bv2 where b= 0.5 N⋅ s2/m2. What is the accelerat

MrMuchimi

The first part of the question is missing and it is:

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

Answer:

a(o) = 18.03 x 10^(-3) m/s^(-2)

Explanation:

First of all, if we make a momentum balance in the direction of motion (the x-direction), we'll discover that the change of the momentum will be equal to the forces acting on the boat.

Hence, there is only the drag force, which acts against the direction of motion as:

d(m·v)/dt = - k·v² (since f_d = - k·v²)

where k = 0.5 Ns²/m²

Now let's simplify the time derivative on the Left, by applying product rule of differentiation and we obtain:

v·dm/dt + m·dv/dt = - k·v²

where dm/dt = 10kg/hr (this is the change of the mass of the boat)

Furthermore, acceleration is the time derivative of time velocity, Thus;

a = dv/dt = - (k·v² + v·dm/dt) / m

Now, for the moment,when the rain starts; since we know all the values on the right hand side, let's solve for the acceleration ;

a(o) = - (ko·vo² + vo·dm/dt) / mo

= -[ (0.5(3)²) + (3x10/3600)/250

(dt = 3600secs because 1hr = 3600 secs)

So a(o) = 18.03 x 10^(-3) m/s^(-2)

3 0

3 years ago

VERY URGENT GIVING BRAINLY​

VERY URGENT GIVING BRAINLY