This diagram shows a baseball motion

A 1.0-kg ball of putty is released from rest and falls vertically 1.5 m until it strikes a hard floor, where it comes to rest in

topjm [15]

<span>We first calculate the velocity of the ball when it hits the ground; this is equal to the square root of the quantity (2*g*d) where g is the acceleration of gravity (9.8 m/s^2) and d is the distance fallen, 1.5m.

So, we get a velocity of sqrt(2*9.8*1.5) = 5.42 m/s.

We can calculate the impulse force applied to the putty ball by using Newton's second law, which states that the applied force is equal to the product of mass and acceleration, where acceleration can be further decomposed as the change in velocity divided by the change in time. Thus, inputting the known values, we have:

F = ma = m(dv/dt) = 1.0*5.42/0.045 = 120.4 newtons.</span>

8 0

3 years ago

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

Terminal speed, v = 6901.07 m/s

Explanation:

It is given that,

Mass of the horizontal bar, m = 30 g = 0.03 kg

Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago

Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the

Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

$q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +delta-p$

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

$0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2} }{2} +0$

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

$0-0=h_{2} -400x10^{3} +\frac{40^{2}-250^{2} }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg$

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

$m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}$

The exit area is:

$m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}$

6 0

3 years ago

The ball in front of you is going to drop one meter. how long will it take to fall?

Pavlova-9 [17]

The ball will take 0.45 seconds to reach the ground

<h3>How will we solve this question?</h3>

We will use Newton’s laws of motion, so

H= $(1/2)gt^2$ taking height= 1m

$1= (1/2) x 9.8 x t^2$

$t=\sqrt{10/49} = 0.45 seconds$

<h3>What are the three laws of motion given by Newton?</h3>

The 3 laws of motion given by Newton are as follows:

1) Unless an unbalanced force acts upon it, an item at rest remains at rest, and an object in motion continues to move in a straight line at a constant pace.

2) An object's acceleration is influenced by its mass and the force being applied.

3) Whenever one thing applies force to another, the second object applies the opposing, equal force to the first.

To know more about Newton's laws of motion visit:

brainly.com/question/27915475

#SPJ4

7 0

2 years ago

which of the following is accurate in describing the placement and classification of iodine? a. iodine is located in period 5, g

Elden [556K]

A. iodine is located in period 5, group 17 and is classified as a nonmetal

Group is the column, period is the raw.

8 0

3 years ago