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irakobra [83]
3 years ago
14

A man pushes a heavy cart with a force exerted of 250 Newtons to keep it moving at a constant velocity. What is the kinetic fric

tional force between the cart and the ground?
Select one:
250 Newtons against the cart's motion
Not enough Information.
246 N against the cart's motion
200 N against the cart's motion
Physics
1 answer:
miv72 [106K]3 years ago
3 0

It can't be less than 250 N or the cart wouldn't move at all. That means there is only 1 answer. It's between not enough info or 250 N. The answer is 250 N. If it was any more, there would be acceleration.

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(3)science who ever gets this right will get a brainlest
lara [203]

Answer:

True.

Explanation:

It also affects your lungs by getting lung disease.

3 0
3 years ago
If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

3 0
3 years ago
What is the IEEE 802.11a center frequency in GHz, and what are its max/min data rates in Mbps? What is the IEEE 802.11b center f
kogti [31]

Explanation:

What is IEEE 802.11?

IEEE 802.11 is a set of WLAN standards for communication developed by the Institute for Electrical and Electronics Engineers (IEEE) and is unarguably most widely used WLAN technology.

Features: IEEE 802.11a

  • The operating frequency band is 5 GHz.
  • The maximum theoretical data rate is 54 Mbps, the typical throughput is around 25 Mbps and minimum data rate is 6 Mbps.
  • It can support 64 users per access point.

Features: IEEE 802.11b

  • The operating frequency band is 2.4 GHz.
  • The maximum theoretical data rate is 11 Mbps but typical throughput is around 6 Mbps and minimum data rate is 1 Mbps.
  • It can support 32 users per access point.

Wireless Coverage IEEE 802.11a Vs IEEE 802.11b:

  • Signal coverage is one of the most important factors among users.
  • The transmission range of IEEE 802.11a is not greater than 100 ft in indoor setting whereas IEEE 802.11b has a superior performance in this regard with transmission range up to 150 ft in indoor setting.
  • The data rate has a direct relation with the access point coverage area, a higher data rate means less coverage area and a lower data rate results in increased coverage.
8 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

7 0
3 years ago
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