What is a scavenger give some examples

Winds tend to rotate in a counter clockwise direction in the ___ (northern or southern) Hemisphere as they move into a low press

lana [24]

Answer:

Winds tend to rotate in a counter clockwise direction in the center of northern and southern hemisphere.

Explanation:

The wind blows clockwise around a high pressure area in the northern hemisphere and the wind blows counter - clockwise around low pressure.

In the northern hemisphere High-pressure systems rotate clockwise direction and in the southern hemisphere low-pressure systems rotate clockwise direction.

8 0

3 years ago

Read 2 more answers

What force, in newtons, must you exert on the balloon with your hands to create a gauge pressure of 62.5 cm H2O, if you squeeze

yulyashka [42]

Answer:54.70 N

Explanation:

Given

Gauge Pressure of $62.5 cm of H_2O$

i.e. $h=62.5 cm =0.625 m$

Effective area $A=51 cm^2$

initial Pressure $= 1 atm=101.325 kPa$

Gauge Pressure $P=\rho gh$

$\rho =density\ of\ water =1000 kg/m^3$

$P_{gauge}=1000\times 9.8\times 0.625=5.937 kPa$

Force creates a pressure of $P_1$ which will be equal to Gauge Pressure

$P_1=\frac{F}{A}$

$P_1=P_{gauge}$

$\frac{F}{A}=5.937 kPa$

$F=5.937\times 51\times 10^{-4}\times 10^3$

$F=30.27 N$

6 0

4 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago

Why is the law of gravity a scientific law

Yuri [45]

An object in motion will stay in motion unless acted upon another force.

Newton used this to prove that gravity existed. Without an unseen force, we could throw a ball and it would go on forever correct? Unless there was something to pull it down, in this case, gravity.

6 0

4 years ago