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nekit [7.7K]
3 years ago
5

The velocity of a projectile at launch has a

Physics
1 answer:
arlik [135]3 years ago
7 0

Answer:

1. 0 vh g

Explanation:

As air resistance is negligible, horizontal speaking, nothing is affecting the velocity. So the horizontal velocity of the projectile stays the same, vh.

As for vertical velocity, since there's always a constant gravitational acceleration acting downward, namely g, the vertical speed will decrease until it reaches the top where it is 0, it then starting to increase in magnitude, downward, due to its gravitational acceleration g.

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The grant that considered the foundation of financial aid is the:
navik [9.2K]

Answer:

I think it is the Federal Pell Grant Program.

Explanation:

4 0
3 years ago
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A body travels a distance of 20m in the 7th second and 24m in the 9th second. How much distance shall it travel in the 15th seco
DaniilM [7]

Answer:

<u>36 m</u>

Explanation:

We can consider this to be an AP.

Then,

  • a₇ = 20
  • a₉ = 24

<u>Subtract a₇ from a₉.</u>

  • a + 8d - a + 6d = 24 - 20
  • 2d = 4
  • d = 2

  • a + 6(2) = 20
  • a = 8

<u>Finding a₁₅</u>

  • a₁₅ = a + 14d
  • a₁₅ = 8 + 14(2)
  • a₁₅ = 8 + 28
  • a₁₅ = <u>36 m</u>
4 0
2 years ago
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A ball rolls down a hill and hits a box. The momentum of the ball decreases. Which happens to the momentum? Select one of the op
Irina-Kira [14]
<span>So we want to know what happens to the momentum of the ball that rolls down hill and hits a box. So we need to use the law of conservation of momentum which states that the momentum must be conserved. It cant be transformed into inertia or mass. It can only be transferred to other object via some interactions like collisions. So it has to be a. transferred to the box and that is the correct answer. </span>
7 0
3 years ago
A 2000-kg truck is being used to lift a 400-kg boulder B that is on a 50-kg pallet A. Knowing the acceleration of the rear-wheel
anzhelika [568]

Answer:

(a) reaction at each front wheel is 5272N (upward)

(b) force between boulder and pallet is 4124N (compression)

Explanation:

Acceleration of the truck a_{t = 1 m/s^{2}  (to the left)

when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,

a_{A} =  0.5 m/s^{2} (upward) , a_{B} =  0.5 m/s^{2} (upward)

Let T be tension in the cable

pallet and boulder: ∑fy = ∑(fy)eff = 2T- (m_{A} + m_{B})g =  (m_{A} + m_{B})a_{B}

                                  = 2T- (400 + 50)*(9.81 m/s^{2}) = (400 + 50)*(0.5 m/s^{2})

                        T = 2320N

Truck:  M_{R} = ∑(M_{R})eff: = -N_{f} (3.4m) + m_{T} (2.0m) - T (0.6m)= m_{T} a_{T} (1.0m)

Nf = (2.0m)(2000 kg)(9.81 m/s^{2} )/3.4m -  (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/s^{2})  = 11541.2N - 409.4N - 588.2N = 10544N

∑fy (upward) = ∑(fy)eff: N_{f} + N_{R} - m_{T}g = 0

                                       10544 + N_{R}  - (2000kg)(9.81 m/s^{2} ) = 0

                N_{R} = 9076N

   ∑fx (to the left) = ∑(fx)eff:  F_{R} - T = m_{T} a_{T}

                                      F_{R} = 2320N + (2000kg)(9.81 m/s^{2} ) = 4320N

(a) reaction at each front wheel:

1/2 N_{f} (upward): 1/2 (10544N) = 5272N (upward)

(b) force between boulder and pallet:

∑fy (upward) = ∑(fy)eff: N_{B} + M_{B}g - m_{B}a_{B}

            N_{B} = (400kg)(9.81 m/s^{2}) + (400kg)(0.5 m/s^{2}) = 4124N (compression)

3 0
4 years ago
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1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del
timofeeve [1]

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

4400 = m*138*363

4400 = 50094m

m = \frac {4400}{50094}

Masa, m = 0.088 kg

7 0
3 years ago
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