All categories

3 years ago

You flip a coin straight up if the coin reaches a high point of 0.25 m above where you released it what was the initial speed?

Physics

1 answer:

Julli [10]3 years ago

6 0

Answer: 2.21 m/s

Explanation:

Given

Coin reaches a height of 0.25 m above the launch point

Suppose u is the initial speed of the coin

Using the equation of motion

$\Rightarrow v^2-u^2=2as\\$

Putting values

$\Rightarrow 0^2-u^2=2\times g\times s=2\times (-9.8)\times 0.25\\\Rightarrow u^2=4.9\\\Rightarrow u=2.21\ m/s$

Thus, the initial velocity is 2.21 m/s

You might be interested in

A wire 1.0 m long experiences a magnetic force of 0.50 N due to a

NNADVOKAT [17]

Answer:

11.50

thenks me later

hehehehe how old are you

6 0

2 years ago

A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0

2 years ago

Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

julsineya [31]

Answer:

$U=50.96J$

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:

U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

U=3×1/4π∈₀×(q²/r)

Substitute given values

$U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\ U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\ U=50.96J$

8 0

3 years ago

What is the slowest speed (in mph) that Mr. P can travel and still jump across the canyon? (I have attached a picture) Hint: The

Alex17521 [72]

might be 140mph, so that is a guess that i just made so plz let me know if im wrong or correct

7 0

2 years ago

The volume of a gas is 605 liters at 27.0°C. The new temperature is -3.0°C. What is the new volume?

nlexa [21]

From p1v1/t1 = p2v2/t2
pressure unchanged ... cancelled out
v1=605 , t1=27C = 300K,
t2=-3C = 270K
***remember temperature must be in Kelvin
we got
605/300 = v2/270
v2 = 545

5 0

2 years ago

Read 2 more answers