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SVEN [57.7K]
2 years ago
13

John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and

picks up the keys to throw them back up to John. If John is 5.33 meters above the sidewalk where Tony is, and Tony throws at 18 m/s at a 40 degree angle above the horizontal, will the keys make it back to John?
Physics
1 answer:
dalvyx [7]2 years ago
8 0

Answer:

the keys will make it back to John.

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

<u>Therefore, the keys will make it back to John.</u>

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Which of the following relationships must be true according to the laws of series and parallel connections? (Select only relatio
tester [92]

Answer:

Explanation:

a False

b True

c True

d False

e False

f False

g True

h False

i False

4 0
2 years ago
The circuit below represents four resistors connected to a 12-volt source. What is the total current in the circuit? 4.0Ω 6.0Ω 1
Tpy6a [65]

Answer:

For series connected resistors I = 0.5A

For parallel connected resistors I = 8.5A

Explanation:

Since the diagram is not available, our solution will be divided into two;

According to ohm's law which states that "the current passing through a,metallic conductor at constant temperature is directly proportional to the potential difference across its ends. Mathematically, E = IRt where;

E is the electromotive force

I is the total current in the circuit

Rt is total equivalent resistance.

Where E = 12volts

Rt can be gotten depending on the arrangements of the resistors which can either be in series or parallel.

If the resistors are in series, their equivalent resistance gives;

Rt = 4.0Ω+6Ω+8Ω+6Ω

Rt = 24Ω

The current I will be;

I = E/Rt = 12/24

I = 0.5A

If the connection is in series, the total current in the circuit will be 0.5A.

For resistance in parallel;

1/Rt = 1/4Ω+1/6Ω+1/8Ω+1/6Ω

1/Rt = 6+4+3+4/24

1/Rt = 17/24

Rt = 24/17Ω

I = E/Rt

I = 12/(24/17)

I = 12×17/24

I = 8.5A

If the connection is in parallel, the total current in the circuit will be 8.5A

8 0
2 years ago
Read 2 more answers
A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with
Lana71 [14]

Answer:

   h = 2 R (1 +μ)

Explanation:

This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the

let's use the mechanical energy conservation agreement

starting point. Lower, just at the curl

       Em₀ = K = ½ m v₁²

final point. Highest point of the curl

        Em_{f} = U = m g y

Find the height y = 2R

      Em₀ = Em_{f}

      ½ m v₁² = m g 2R

       v₁ = √ 4 gR

Any speed greater than this the body remains in the loop.

In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law

X axis

    -fr = m a                      (1)

Y Axis  

      N - W = 0

      N = mg

the friction force has the formula

     fr = μ  N

     fr = μ m g

    we substitute 1

    - μ mg = m a

     a = - μ g

having the acceleration, we can use the kinematic relations

    v² = v₀² - 2 a x

    v₀² = v² + 2 a x

the length of this zone is x = 2R

    let's calculate

     v₀ = √ (4 gR + 2 μ g 2R)

     v₀ = √4gR( 1 + μ)

this is the speed so you must reach the area with fricticon

finally have the third part we use energy conservation

starting point. Highest on the ramp without rubbing

     Em₀ = U = m g h

final point. Just before reaching the area with rubbing

     Em_{f} = K = ½ m v₀²

      Em₀ = Em_{f}

     mgh = ½ m 4gR(1 + μ)

       h = ½ 4R (1+ μ)

       h = 2 R (1 +μ)

7 0
2 years ago
A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "
Alexus [3.1K]

Answer:

Explanation:

The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.

I amu = 931 Mev

2.2 Mev = 2.2 / 931 amu

= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷

= .00395 x 10⁻²⁷

The mass of the deuteron  =( 1.6726 + 1.6749 +  .00395)x 10⁻²⁷ kg

= 3.35145 x 10⁻²⁷ kg

b ) Momentum of gamma ray

= h / λ ( h is plank's constant and λ is wavelength of gamma ray )

= hυ / υλ       (  υ is frequency of gamma ray )

= E / c  ( E is energy of photon and c is velocity o light )

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J / 3 x 10⁸

= 1.173 x 10⁻²¹ Kg m /s

This will be the momentum of deuteron also

Kinetic energy

= p² / 2m ( p is momentum and m is mass of deuteron )

= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)

= 1.376 x ⁻¹⁵ J

Energy of gamma ray

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J

= 3.52 x 10⁻¹³ J

So kinetic energy of deuteron is smaller than energy of gamma ray photon .

5 0
2 years ago
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
2 years ago
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