All categories

3 years ago

Strong evidence for the existence of dark matter comes from observations of

Physics

1 answer:

adoni [48]3 years ago

5 0

Answer:

NASA's Chandra X-ray Observatory and other telescopes such as the Hubble telescope.

Explanation:

Dark matter and normal matter have been wrenched apart by the tremendous collision of two large clusters of galaxies.

The above observations have provided the strongest evidence yet that most of the matter in the universe is dark.

You might be interested in

Are momentum and kinetic energy conserved during an inelastic collision?

nevsk [136]

Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision though. And that is because it is converted to another form of energy

8 0

3 years ago

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Now in the given problem

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

u = 51.5 m/s, for maximum range θ = 45⁰

So maximum distance reached = $\frac{51.5^2sin(2*45)}{9.81}=270.36 meter$

So it will go a maximum of 270.36 meter.

5 0

3 years ago

A charge of 2 c is at the origin. when charge q is placed at 2 m along the positive x axis, the electric field at 2 m along the

docker41 [41]

The value of Q will be -8 C.

In the presence of an electric or magnetic field, matter experiences a force due to its electric charge.

A moving electric charge generates a magnetic field, and an electric charge has an accompanying electric field.

The information provided in the issue is;

The separation between and is 2m.

An origin charge equals +2 C

The electric fields are identical in magnitude but are facing in different directions. As a result, the following relationship can be used

Q/16=1/2

The value of Q will be -8 C.

Learn more about electric charge here

brainly.com/question/8163163

#4174

4 0

2 years ago

You push against a steamer trunk with a force of 800 n at an angle alpha with the horizontal . the trunk is on a flat floor and

galina1969 [7]

The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha

6 0

3 years ago

It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago