What is the acceleration of a 10 kg mass pushed by a 5 N force

A train moves at constant velocity of 50km/h. how far will it move in 0.5h?

Kobotan [32]

The formula for getting the distance will be distance = speed x time
                                                                               D = S x T
   speed or velocity = 50km/h
   time = 0.5 h
the equation will be done directly because it's already in it's SI units
            distance = 50km/h x 0.5h
            hour cancels hour and the equation remains = 50km x 0.5
                               Ans = 25 km
the train will move 25 km far in 0.5h

7 0

3 years ago

One event occurs at the origin at t equal to zero, and a second events occurs at the point x=5m along the x-axis at time with ct

Anastasy [175]

Answer:

The separation will be spacelike.

The first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

Explanation:

We can define the separation between two events (using the + - - - signature) as :

$(\Delta s )^2 = (ct_2 - c t_1 )^2 - (x_2 - x_1)^2$

where the separation will be lightlike if is equal to zero, timelike if is positive and spacelike if is negative.

For our problem

$c t_1 = 0$

$x_1 = 0$

$ct_2 = 4 \ m$

$x_2 = 5 \ m$

$(\Delta s )^2 = (4 \ m - 0 )^2 - ( 5 \ m - 0)^2$

$(\Delta s )^2 = (4 \ m )^2 - ( 5 \ m 0)^2$

$(\Delta s )^2 = 16 \ m^ 2 - 25 \ m^2$

$(\Delta s )^2 = - 9\ m^2$

So the separation will be spacelike, and the first event can't cause the second event, as there exist an frame of reference in which both happens at the same time, in different positions, so, if there were causally connected, it will imply an instant connection, this is faster than light.

8 0

3 years ago

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . Suppose that

RUDIKE [14]

Answer:

0.0241875 m

Explanation:

$m_1$ = Mass of quarterback = 80 kg

$m_2$ = Mass of football = 0.43 kg

$v_1$ = Velocity of quarterback

$v_2$ = Velocity of football = 15 m/s

Time taken = 0.3 seconds

In this system as the linear momentum is conserved

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{0.43\times 15}{80}\\\Rightarrow v_1=0.080625\ m/s$

Assuming this velocity is constant

$Distance=Velocity\times Time\\\Rightarrow Distance=0.080625\times 0.3\\\Rightarrow Distance=0.0241875\ m$

The distance the quarterback will move in the horizontal direction is 0.0241875 m

4 0

3 years ago

Which best summarizes the scientific process that led to our current understanding of DNA? Please help

solniwko [45]

The answer is B.)

This is because for years scientists have build up and found discoveries that led to recent discoveries brought by previous scientists.

Hope this helps

7 0

3 years ago

Read 2 more answers

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$ ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$ ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m

7 0

3 years ago