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What is the force produced by a 10kg mass accelerating at 2.5M's?
Wiki User
∙ 11y ago
A mass doesn't produce a force by accelerating. It needs a force to make it accelerate.
The force it needs is (mass) x (acceleration).
In order to accelerate a 10 kg mass at 2.5 meters per sec2, you need a force of 25 newtons.
Wiki User
∙ 11y ago
Wiki User
∙ 12y agoThe question can't be answered because it doesn't give a consistent set of
information. There's no such thing as 2 kilograms of force, and that phrase
has no meaning. We're left not knowing the magnitude of the force acting
on the 10 kg body. The question uses "kg" to express both a mass and
a force. You can get away with that in the English/Imperial/customary
system with 'pounds', at least for a while, but not in the metric system.
If you say that a body with 10 kg of mass weighs 10 kg, I'll say OK, but then
I'll ask you what the same body weighs on the moon. The answer is that the
body with 10 kg of mass weighs 1.6 kg on the moon, and it's obvious that
you have painted yourself into a corner from which there is no escape.
Anonymous
answer in 2 newton
Anonymous
answer is 2 newton
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Mass is 145g and velocity is 25ms Find the answer using kinetic energy formula?
Kinetic energy is 0.45 joules using the formula Ek = ½ x mv2
What is the mass of a baseball thrown at 25ms resulting in a momentum of 10kgms?
Just use the definition of momentum, as mass x velocity. In this case, you need to divide the momentum by the velocity, to get the mass.
A car moving at 10ms is accelerated at a rate of 5ms2 over a 4 second period What would the speed be after 3 seconds?
I presume that you mean 'An object travelling initally at 10m/s, accelerates at 4m/s2 for 8s; what is its final speed? If it is accelerating at 4 m/s2, it gains 4m/s every s v= u + at where v is the final speed, u is the initial speed, a is the acceleration and t is the time v = 10 + 4(8)= 42 m/s
A stone projected horizontally from the top of a vertical wall with a velocity of 25ms hits the horizontal ground at a point 60m from the base of wall calculate the time of flight of the stone g10ms?
im guessing this is homework huh?
A stone is allowed to fall from the top of a tower 100m high and at the same time other stone is projected vertically upwards with a velocity of 25ms Calculate when and where the two stones will meet?
The altitude of the first stone above the ground is [ H = 100 - 1/2 G T2 ].The altitude of the second stone is [ H = 25T - 1/2 G T2 ].The stones meet when their altitudes are the same.100 + 1/2 G T2 = 25T - 1/2 G T2Add 1/2 G T2 to each side of the equation:25 T + GT^2 -100 = 0They meet when T = 2.15 seconds after the drop-and-tossPlug that into the expression for the position of the dropped stone:100 - 1/2 G T2 = 100 - (4.9) (16) = 100 - 78.4 = 21.6 meters from the ground.
When mass is 5 kg and accelaration is 25ms-2 what will be the force?
When Using Newtons Laws you get: F=m*a=5 kg*25 m/s2=125 N
Which glass pitbull using eye glass in international love?
I think it's prada 25ms
Mass is 145g and velocity is 25ms Find the answer using kinetic energy formula?
Kinetic energy is 0.45 joules using the formula Ek = ½ x mv2
What is the mass of a baseball thrown at 25ms resulting in a momentum of 10kgms?
Just use the definition of momentum, as mass x velocity. In this case, you need to divide the momentum by the velocity, to get the mass.
A car moving at 10ms is accelerated at a rate of 5ms2 over a 4 second period What would the speed be after 3 seconds?
I presume that you mean 'An object travelling initally at 10m/s, accelerates at 4m/s2 for 8s; what is its final speed? If it is accelerating at 4 m/s2, it gains 4m/s every s v= u + at where v is the final speed, u is the initial speed, a is the acceleration and t is the time v = 10 + 4(8)= 42 m/s
A stone projected horizontally from the top of a vertical wall with a velocity of 25ms hits the horizontal ground at a point 60m from the base of wall calculate the time of flight of the stone g10ms?
im guessing this is homework huh?
How do you calculate mean latency?
To find the mean of anything, including latency, add all the values together and divide the result of that by the total number of input values. So, say you had this: 24ms 57ms 25ms 45ms The total of these numbers is 151 There were 4 values when we started so we divide our total by 4 151/4 = 37.75 So our mean latency is 37.75ms Hope this helps!
How many milliseconds after the zero crossing of the AC voltage reaches of 1kHz and 10kHz for the first time its maximum value?
1khz you get 2000 zero crossings per sec or .5 ms between zeros depending on which zero crossing you pick it may be .25ms or .75ms to the max or min for ten khz .05ms between zeros .025ms or .075ms
What is ip address of a website?
A website name is resolved to the IP address by a DNS (Domain Name Server) allowing websites to be navigated to via their familiar domain name (such as "Google.com") rather than to a hard-to-remember IP address. IP addresses used by networks and the World-Wide Web to uniquely identify a domain (such as a computer or website). Websites are identified uniquely using an IP address, although physically there may be several servers handling the website. To find out the IP address of a website, open a command prompt and type "ping <website URL>"- the IP address of the website will be resolved and displayed for the given domain. Here is an example: C:\>ping google.com Pinging google.com [74.125.91.106] with 32 bytes of data: Reply from 74.125.91.106: bytes=32 time=27ms TTL=49 Reply from 74.125.91.106: bytes=32 time=27ms TTL=49 Reply from 74.125.91.106: bytes=32 time=25ms TTL=49 Reply from 74.125.91.106: bytes=32 time=29ms TTL=49 Ping statistics for 74.125.91.106: Packets: Sent = 4, Received = 4, Lost = 0 (0% loss), Approximate round trip times in milli-seconds: Minimum = 25ms, Maximum = 29ms, Average = 27ms The IP address in this example (for google.com) is 74.125.91.106
A stone is allowed to fall from the top of a tower 100m high and at the same time other stone is projected vertically upwards with a velocity of 25ms Calculate when and where the two stones will meet?
The altitude of the first stone above the ground is [ H = 100 - 1/2 G T2 ].The altitude of the second stone is [ H = 25T - 1/2 G T2 ].The stones meet when their altitudes are the same.100 + 1/2 G T2 = 25T - 1/2 G T2Add 1/2 G T2 to each side of the equation:25 T + GT^2 -100 = 0They meet when T = 2.15 seconds after the drop-and-tossPlug that into the expression for the position of the dropped stone:100 - 1/2 G T2 = 100 - (4.9) (16) = 100 - 78.4 = 21.6 meters from the ground.
A stone is dropped from the top of a tower 100m in height and at the same time another stone is projected vertically upwards from the ground with a velocity of 25ms find where and when the two stone?
The position of an object in a gravity field is given by: S = 1/2 * a *t ^2 + v * t + x So making the assumption that both are subject to Earth's gravity, a = - 9.81 m/s^2 (negative acting down). Assuming the top stone starts at rest, V_top = 0 m/s, V_bottom = 25 m/s. Taking the problem statement, x_top = 100 m and x_bottom = 0 m. If you want the time when they impact one another, set the two position equations equal and solve for t: 1/2* - 9.81 * t ^2 + 0 + 100 = 1/2 * -9.81 * t^2 + 25 * t + 0 So the acceleration terms cancel, and t = 4 seconds. To check, solve both position equations at t = 4 s. S = 1/2 * -9.81 * 4 ^2 + 0 + 100 = 21.52 m S = 1/2 * -9.81 * 4 ^ 2 + 25 * 4 + 0 = 21.52 m So there you go. The time is 4 seconds, the position is 21.52 m above the ground.
What Asynchronous voice band modems?
Synchronous modems operates in the audio domain, at rates up to 28800 bps in audio lines, used in telephones systems (using synchronous data). The usual modulation methods are the phase modulation and integrated phase and amplitude (at higher rates than 4800 bps).In synchronous modems, equalizers are used, in order to offset the misfit of the telephone lines. These equalizers are inserted in addition to the equalizers, that sometimes already exist in the telephone lines.These equalizers can be classified into three main groups:Fixed/statistical equalizer - these equalizers offset the signal according to the average of the known attenuation in each frequency. Tuning the equalizer is sometimes done in the factory and stays fixed, usually they are used to operate at low rates in a dial up line.Manually adjusted equalizer - these equalizers can be tuned to optimal performance to a given line. These equalizers should be re-tuned when the line is replaced and periodically. Specially, it should be tuned frequently when the line is of a low quality and it's parameters are changed frequently. Tuning is done using a button inside the modem (or on the external board).Automatic equalizer - these equalizers are tuned automatically when the connection is established. Depending on the line quality in a specific moment, in a process of about 15ms to 25ms, after the first tuning, the equalizer samples the line continually and adjusts itself to the changed conditions, so the modem operates at each moment under optimal conditions. The fitness process operates, in some modems, at rates of 2400 times in a second.Synchronous modems operate in the same manner asynchronous modems. However, synchronous modems operates at higher rates and since the requirements to transmit at these rates is increasing, most of the innovations are implemented for synchronous modems.In synchronous modems the channel can be split for several consumers at various speeds. Modems who have this ability are called SSM - Split System Modem. These modems can use a simple split or a split using multipoint connection.Synchronous data is accompanied by a clock signal. Synchronous data is almost always grouped in blocks, and it is the responsibility of the data source to assemble those blocks with framing codes and any extra bits needed for error detecting and/or correcting according to one of many different protocols (BISYNC, SDLC, HDLC, etc.). The data source and destination expect the modem to be transparent to this type of data, conversely, the modem can ignore the blocking of the data.
