All categories

3 years ago

Which of the following is the brightness of a star as we see if from Earth?

Physics

1 answer:

My name is Ann [436]3 years ago

8 0

I think that it is apparent magnitude

You might be interested in

What is an example of 'Centre of Gravity'?

Lera25 [3.4K]

Water bottle and a paper towel role

5 0

3 years ago

What is the total resistance of this second circuit show work

Leya [2.2K]

$R_\Sigma = R_1 + R_2 + R_3 = 34 \ \Omega.$

4 0

3 years ago

Read 2 more answers

A deer is running from a mountain lion when it encounters a fence that is 1.50 m high. Seeing the fence, the deer jumps, leaving

Novosadov [1.4K]

Answer:

Explanation:

Given

Initial Velocity (u)=13 m/s

angle $=29^{\circ}$

distance between fence and deer=2.5 m

We consider deer jump similar to projectile motion

equation of trajectory

$y=xtan\theta -\frac{gx^2}{2u^2(cos\theta )^2}$

$y=2.5tan(29)-\frac{9.8\times 2.5^2}{2\times 13^2\times (cos29)^2}$

$y=1.385-0.2368=1.148 m$

Thus deer will cross the fence with an difference in its jump and fence

1.5-1.148=0.351 m

$h_{max}=\frac{u^2sin^2\theta }{2g}$

$h_{max}=\frac{13^2\times (sin29)^2}{2\times 9.8}$

$h_{max}=2.02 m$

so deer rises during when it is over fence

8 0

4 years ago

A 2.75 kg block is pulled across a flat, frictionless floor with a 5.11 N force directed 53.8° above horizontal. What is the tot

Softa [21]

Answer:

F = 3.01 N

Explanation:

Given that,

Mass of a block, m = 2.75 kg

Force applied to the block, F = 5.11 N

It is directed 53.8° above horizontal.

We need to find the total force acting on the block. The force acting on it is given by :

$F=F\cos\theta\\\\F=5.11\times \cos53.8\\F=3.01\ N$

So, 3.01 N of force is acting on the block.

4 0

3 years ago

After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.40 rad/s and it rotated 12.3 re

zaharov [31]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of angular motion.

By definition, acceleration can be expressed as the change in angular velocity squared over a given period of distance traveled.

$\alpha = \frac{\omega^2}{2\theta}$

where,

$\omega =$ Angular velocity

$\theta =$ Angular displacement.

In turn, as a function of time, we can represent it as,

$\alpha = \frac{\omega}{t}$

For our case we have to,

$\omega = 5.4rad/s$

$\theta = 12.3rev = 12.3rev(\frac{2\pi rad}{1rev})=24.6\pi rad$

PART A) In the case of angular acceleration we have to,

$\alpha = \frac{\omega^2}{2\theta}$

$\alpha = \frac{(5.4)^2}{2*24.6\pi}$

$\alpha = 0.1886rad/s^2$

PART B) Through the definition of angular acceleration as a function of time we can calculate it,

$\alpha = \frac{\omega}{t}$

$t = \frac{\omega}{\alpha}$

$t = \frac{5.4}{0.1886}$

$t = 28.63s$

3 0

4 years ago