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exis [7]
3 years ago
7

A student throws a ball upward with a velocity of 35 m/s. What is the acceleration of the ball as it rises to the top of its arc

?
Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

The acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

Explanation:

Let suppose that maximum height of the arc is so small in comparison with the radius of the Earth.

Since the ball is launched upwards, then the ball experiments a free-fall motion, that is, an uniform accelerated motion in which the element is accelerated by gravity. Then, the acceleration experimented by the motion remains constant at every instant and position.

Besides, the gravitational acceleration in the Earth and, in consequence, the acceleration of the ball as it rises to the top of its arc equals 9.807 meters per square second.

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lood flows through a section of a horizontal artery that is partially blocked by a deposit along the artery wall. As a hemoglobi
uysha [10]

Answer:

P_1 - P_2 =219.62\ Pa

Explanation:

given,

density of blood = 1060 kg/m³

v₂ = 0.800 m/s

v₁ = 0.475 m/s

change in pressure calculation

using Bernoulli's equation

P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2

both are at same level

P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)

P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)

P_1 - P_2 =219.62\ Pa

the change in pressure is equal to 219.62 Pa

5 0
3 years ago
A tungsten wire has resistance R at 20°C. A second tungsten wire at 20°C has twice the length and half the cross-sectional area
Bad White [126]
The resistance is 4 times the resistance of the first wire. the formula is R = p*l/A with p being resistivity, l length and A area. So if you double length and half area, which botv result in more resistance, you get p*2/0.5 or 4 (p can be abandoned because it is the same. We take standard length and area as 1)
6 0
3 years ago
Question 2
Yakvenalex [24]

Answer:

B

Explanation:

6 0
3 years ago
Energy from solar radiation may be ________ or taken in by a surface or an object.
tester [92]

Answer:

Absorbed

Explanation:

I hope this helps you

7 0
3 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
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