Statement: Baseball is not a dangerous sport unless either a bat breaks or a player is hit with a ball.Key: B = A bat breaks.
D = Baseball is a dangerous sport.
P = A player is hit with a ball.

Answers

Answer 1

Answer:

Answer:

Explanation:

If the bat break it will get replaced and you will get another bat,but if u get hit with the ball it's dangerous in many ways

Answer 2

Answer:

Final answer:

The question seeks to apply Boolean logic, specifically the 'OR' function, to the statement about baseball. In Boolean terms, 'Baseball is a dangerous sport (D)' is true if either 'A bat breaks (B)' or 'A player is hit with a ball (P)' is true.

Explanation:

The subject matter of this question is Boolean logic, which is a subfield of mathematics. In the provided statement, the sport of baseball (D) is deemed dangerous if either a bat breaks (B) or a player is hit with a ball (P). In terms of Boolean logic, this can be symbolized as 'D = B ∨ P', where ∨ signifies the logical operation 'OR'. Therefore, if either B or P is true (meaning a bat breaks or a player is hit), then D (Baseball is a dangerous sport) is consequentially deemed true.

Learn more about Boolean logic here:

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The time taken by a mass projected verticallyupwards to reach the maximum height (with air
resistance not neglected) is 10 sec. The time of
descent of the mass from the same height will be

Answers

Answer:

10s

Explanation:

The time to get to the maximum would be the same as the time to get down to the maximum unless somehow gravity’s changes during the duration it goes up to and from maximum height.

Please show steps as to how to solve this problem
Thank you!

Answers

Answer:

Torques must balance

F1 * X1 = F2 * Y2

or M1 g X1 = M2 g X2

X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7

X2 = 22.4 cm

Torque = F1 * X2 =

62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2

Normally x cross y will be out of the page

r X F for F1 will be into the page so the torque must be negative

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

$P_r=(IA)/(c)$

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

$P=140W=140(J)/(s)\n0.4*140J=56J$

56J is the electromagnetic energy.

The area of the bulb is

$A_b=4\pi r^2=4\pi (14m)^2=2463m^2$

The radiation pressure is

$P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa$

hope this helps!!

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Answers

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.

Answers

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is = 2.93 × 109 W/m2. What is the rms value of (a) the electric field and (b) the magnetic field in the electromagnetic wave emitted by the laser?

Answers

A. The rms value of electric field be "1.05 × 10⁶ N/C".

B. The rms value of magnetic field will be "3.5 × 10⁻³ T".

Magnetic and Electric field

According to the question,

Intensity of the wave, S = 2.93 × 10⁹ W/m²

Free space permittivity, $\epsilon_0$ = 8.86 × 10⁻¹²

Speed of light, c = 3 × 10⁸

A. We know that,

The rms value of electric field,

→ $E_(rms)$ = $\sqrt{(S)/(\epsilon_0 c) }$

By substituting the values,

= $\sqrt{(2.93* 10^9)/((8.85* 10^(-12))(3* 10^8)) }$

= 1.05 × 10⁶ N/C

and,

B. We know that,

The rms value of magnetic field,

→ $B_(rms)$ = $(E_(rms))/(c)$

By substituting the values,

= $(1.05* 10^6)/(3* 10^8)$

= 3.5 × 10⁻³ T

Thus the above response is appropriate.

Find out more information about magnetic field here:

brainly.com/question/25801845

To solve this problem, it is necessary to apply the concepts related to the electric field according to the intensity of the wave, the permittivity constant in free space and the speed of light.

As well as the expression of the rms of the magnetic field as a function of the electric field and the speed of light.

PART A) The expression for the rms of electric field is

$E_(rms) = \sqrt{(S)/(\epsilon_0 c)}$