Scientist who discovered that the nucleus contains neutrons in addition to protons.

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height

butalik [34]

Answer:

a) I = -2257.6 Kg*m/s

b) F = -451,520N

Explanation:

part a.

we know that:

I = $P_f-P_i$

where I is the impulse, $P_f$ the final momentum and $P_i$ the initial momentum.

so:

I = $MV_f-MV_i$

where M is the mass, $V_f$ the final velocity and $V_i$ the initial velocity.

Therefore, we have to find the initial velocity or the velocity of the passenger just before the collition.

now, we will use the law of the conservation of energy:

$E_i=E_f$

so:

mgh = $\frac{1}{2}MV_i^2$

where g is the gravity and h the altitude. So, replacing values, we get:

(85kg)(9.8m/s^2)(36m)= $\frac{1}{2}(85kg)V_i^2$

solving for $V_i$ :

$V_i = 26.56m/s$

Then, replacing in the initial equation:

I = $MV_f-MV_i$

I = $(85kg)(0m/s)-(85kg)(26.56m/s)$

I = -2257.6 Kg*m/s

Then, the impulse is -2257.6 Kg*m/s, it is negative because it is upwards.

part b.

we know that:

Ft = I

where F is the average force, t is the time and I is the impulse. So, replacing values, we get:

F(0,005s) = -2257.6 Kg*m/s

solving for F:

F = -451520N

Finally, the force is -451,520N, it is negative because it is upwards.

3 0

3 years ago

A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loo

m_a_m_a [10]

Answer:

The induced emf is $\epsilon = 0.1041 \ V$

Explanation:

From the question we are told that

The radius of the circular loop is $r = 9.50 \ cm = 0.095 \ m$

The intensity of the wave is $I = 0.0215 \ W/m^2$

The wavelength is $\lambda = 6.90\ m$

Generally the intensity is mathematically represented as

$I = \frac{ c * B^2 }{ 2 * \mu_o }$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4 \pi *10^{-7} N/A^2$

B is the magnetic field which can be mathematically represented from the equation as

$B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }$

substituting values

$B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }$

$B = 1.342 *10^{-8} \ T$

The area is mathematically represented as

$A = \pi r^2$

substituting values

$A = 3.142 * (0.095)^2$

$A = 0.0284$

The angular velocity is mathematically represented as

$w = 2 * \pi * \frac{c}{\lambda }$

substituting values

$w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }$

$w = 2.732 *10^{8} rad \ s^{-1}$

Generally the induced emf is mathematically represented as

$\epsilon = N * B * A * w * sin (wt )$

At maximum induced emf $sin (wt) = 1$

So

$\epsilon = N * B * A * w$

substituting values

$\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}$

$\epsilon = 0.1041 \ V$

7 0

3 years ago

Use the graph below to answer the following question: What is happening to the object's velocity?

hjlf

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

The y-axis represent displacement.
The x-axis depicts time
Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

4 0

3 years ago

A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te

Nitella [24]

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

6 0

3 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

3 years ago