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Mumz [18]
3 years ago
7

What is the average acceleration of a rocket that travels 400m in 5s if it starts from rest?​

Physics
1 answer:
Pavel [41]3 years ago
4 0

Answer:

<h2>32m/s^2</h2>

Explanation:

We want to find the acceleration based on the given data

Given

distance s=400m

time t= 5s

u= 0m/s since it started from rest

We apply the following expression

s=ut+1/2at^2

substituting we have

400=0*5+1/2*a(5)^2

400=25a/2

cross multiply

25a=400*2

25a=800

divide both sides by 25 we have

a=800/25

a=32m/s^2

The acceleration is 32m/s^2

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An astronaut's training centrifuge has a radius of 4.0m. If it goes round once every 2.5s, calculate the velocity of the end of
alexandr1967 [171]

Answer: 10.048m/s

Explanation:

We know that the radius is r = 4.0m

And for rotating things, the tangential velocity (or the velocity of the end of the centrifuge arm) can be calculated as:

v = r*w

The period is T = 2.5s,

This means that if sin(w*t) describes this situation, we have that:

sin(w*t) = sin(w*(t + T))

and we know that:

sin(w*0) = 0

sin(w*(0 + T)) = Sin(w*T) = 0

this means that w*T = 2*pi

w = 2*pi/T = 2*pi/2.5s = (2*3.14)/2.5 s = 2.512 hz

Then the velocity can be calculated as

v = r*w = 4.0m*2.512hz = 10.048m/s

6 0
3 years ago
A cylindrical region of radius R contains a uniform magnetic field parallel to its axis. The field is zero outside the cylinder.
PtichkaEL [24]

The magnitude of the induced electric field is (RdB/dt)/4

The induced electric field is gotten from

-∫E.dl = dФ/dt where E = induced electric field, dl = path length vector, Ф = magnetic flux through cylindrical region = AB where A = area of magnetic flux = πR² where R = radius of cylindrical region and B = magnetic field.

So, -∫E.dl = dФ/dt

-∫E.dl = dAB/dt

-∫Edlcos0 = AdB/dt (where E.dl = Edlcos0 = Edl since E and dl are parallel to each other.)

So -∫Edl = πR²dB/dt

-E∫dl = πR²dB/dt (∫dl = 2πr since the integral is the circumference of the path)

-E(2πr) = πR²dB/dt (we integrate dl from r = 0 to 2R)

-E2π(2R - 0) = πR²dB/dt

-E4πR= πR²dB/dt

E = πR²dB/dt ÷ 4πR

E = -(RdB/dt)/4

So, the magnitude of the induced electric field is (RdB/dt)/4

Learn more about induced electric field here:

brainly.com/question/15730392

3 0
2 years ago
Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) A baseball is throw
wel

Answer:8.1 m

Explanation:

Given

ball is launched from height of 3 m

initial velocity u=10 m/s

considering the ball is thrown vertically upward

Using v^2-u^2=2 as

where,

u=initial velocity

v=final Velocity

a=acceleration

s=distance

At maximum height final velocity will be zero

v^2-u^2=2 gs

0-10^2=2(-9.8)\cdot h

h=5.10 m

Therefore maximum height w.r.t ground is 3+5.10=8.10 m

8 0
3 years ago
Cross-training employees is only necessary in the genetic science industry.<br> O True<br> O False
zhannawk [14.2K]

Answer: False

Explanation:  

8 0
3 years ago
Read 2 more answers
Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of
Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{high} = 15.999\,kPa

p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}

p_{low} = 10.666\,kPa

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}

p_{high} = 2.320\,psi

p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}

p_{low} = 1.547\,psi

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}

h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg} (Eq. 1)

Where:

\rho_{w}, \rho_{Hg} - Densities of water and mercury, measured in kilograms per cubic meter.

h_{w}, h_{Hg} - Heights of water and mercury columns, measured in meters.

If we know that \rho_{w} = 1000\,\frac{kg}{m^{3}}, \rho_{Hg} = 13600\,\frac{kg}{m^{3}}, h_{Hg, high} = 0.120\,m and h_{Hg, low} = 0.080\,m, then we get that:

h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m

h_{w, high} = 1.632\,m

h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m

h_{w, low} = 1.088\,m

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0
3 years ago
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