Answer:
For this you would use planck's equation.

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14 x 6.63 x 10^-34

= 3.45 x 10^-19 Joules

E = hv, where v = the frequency and h = planck's constant.

So E = 5.20 x10^14 x 6.63 x 10^-34

= 3.45 x 10^-19 Joules

____ can be calculated if you know the distance that an object travels in one unit of time. A.motionB.meterC.RateD.SpeedE.velocityF.slopeG.refrence point PLS HELP NOW !!!

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

A soccer ball is kicked straight upwards with an initial vertical speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m 8, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. How long does it take the ball to have a downwards speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?

The concentration of Biochemical Oxygen Demand (BOD) in a river just downstream of a wastewater treatment plant’s effluent pipe is 75 mg/L. If the BOD is destroyed through a first-order reaction with a rate constant equal to 0.05/day, what is the BOD concentration 50 km downstream? The velocity of the river is 15 km/day.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Tonya picks up a leaf from the ground and holds it at arm’s length. She lets go, and the leaf falls to the ground. What force pulled the leaf to the ground?

A soccer ball is kicked straight upwards with an initial vertical speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m 8, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. How long does it take the ball to have a downwards speed of 4.0\,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?

You're going to use the constant acceleration motion equation for velocity and displacement:

(V)final² = (V)initial²+2a(dx)

Given:

a=0.500m/s²

dx=6.32 m

(V)intial=0m

(V)final= UNKNOWN

(V)final= 2.51396m/s

Part 1: here are the answers in order

5

2

1

3

4

5

2

1

3

4

The **range** of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

To solve this problem, we need to make use of the concept of** projectile motion** in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

#SPJ3

**Answer:**

**Explanation:**

In this case, power is the rate of transferring heat per unit time:

The heat is given by the formula of the latent heat of fusion, since the ice is melting.

Here m is the ice's mass and is the heat of fusion of ice. Recall that one day has 86400 seconds. Replacing (2) in (1) and solving:

2. It is moving to the right with a net force of 10 N.

3. It is in dynamic equilibrium with a net force of 0 N.

4. It is in static equilibrium with a net force of 0 N.

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting **free-body diagram**. (option 3)

**A free-body diagram **is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are **two forces **acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in **dynamic equilibrium**.

Learn about **free-body diagram** here brainly.com/question/10148657

#SPJ3

**Answer:**

4. It is in static equilibrium with a net force of 0 N.

**Explanation:**

Just got it right :)

Answer:

Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²

Explanation:

Please see attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.

The question is about understanding **kinematics **in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two **components,** one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the** Pythagorean theorem**: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the** acceleration **is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².

#SPJ3