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KengaRu [80]
3 years ago
7

What's a definition of caught​

Physics
1 answer:
disa [49]3 years ago
6 0

Answer:

past and past participle of catch.

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A projectile is fired horizontally from a height of 78.4 m at a speed of 300 m/sec. How far did it travel horizontally before hi
Mice21 [21]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g

U is the speed = 300m/s

H is the maximum height = 78.4m

g is the acceleration due to gravity = 9.8m/s²

Substitute into the fromula;

R = 300√2(78.4)/9.8

R = 300 √(16)

R = 300*4

R = 1200m

Hence the projectile travelled 1200m before hitting the ground

3 0
3 years ago
A current of 0.2 A flows through a conductor for 5minutes. How much charge would have passed through the conductor?​
Montano1993 [528]

As we know,

electric \:  \: current =  \dfrac{charge}{time}

so, let's solve for charge (q) :

time = 5 minutes = 5 × 60 seconds = 300 seconds.

  • 0.2 =  \dfrac{q}{300}

  • q = 300 \times 0.2

  • q = 60

hence, the charge = 60 coulombs (C)

4 0
2 years ago
Ask Your Teacher The height h in feet reached by a dolphin t seconds after breaking the surface of the water is given by h = −16
kirza4 [7]

Answer:

1 second

Explanation:

h = −16t² + 32t

When, h = 16

16 = −16t² + 32t

Divide each of the numbers by 16

1 = -1t² + 2t

Rearrange the equation

1t²-2t+1 = 0

Solving by the quadratic formula, we get

t=\frac{-(-2)\pm \sqrt{(-2)^2-4\times 1\times 1}}{2\times 1}\\\Rightarrow t=\frac{2\pm 0}{2}\\\Rightarrow t=1\ s

So, time taken by the dolphin to jump out of the water and touch the trainer's hand is 1 second.

3 0
3 years ago
A solenoid has 450 loops each of radius 0.0254 m. The field increases from 0 T to 3.00 T in 1.55 s. What is the EMF generated in
allochka39001 [22]

Answer:

0.175 second

Explanation:

i hope it helps

8 0
2 years ago
A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that
dimaraw [331]

Answer:

0.022m or 2.2cm

Expxlanation:

Step 1:

Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

6 0
3 years ago
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