All categories

3 years ago

If 400 Pa pressure is applied to an area of 55 m2, which is the resulting force

Physics

2 answers:

Novay_Z [31]3 years ago

8 0

You have to do 400.Pa×55m2
the answer is 22000

madam [21]3 years ago

5 0

Force = Pressure × Area

Pressure = 400Pa
Area = 55 m²

Force = 400 × 55 = 22000 N

You might be interested in

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

v is the speed

r is the radius

Now,

$a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2$

In g units:

$a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g$

7 0

2 years ago

Hooke's law describes a certain light spring of unstretched length 31.8 cm. when one end is attached to the top of a doorframe a

lubasha [3.4K]

Answer:

Explanation:

extension in the spring = 40.4 - 31.8 = 8.6 cm = 8.6 x 10⁻² m .

kx = mg

k is spring constant , x is extension , m is mass

k x 8.6 x 10⁻² = 7.52 x 9.8

k = 856.93 N/m

= 857 x 10⁻³ KN /m

b ) Both side is pulled by force of 188 N .

Tension in spring = 188N

kx = T

856.93 x = 188

x = .219.38 m

= 21.938 cm

= 21.9 cm .

length of spring = 31.8 + 21.9

= 53.7 cm .

6 0

2 years ago

Two persons manage to push a motorcar of mass 1200 kg at uniform velocity along a level road, The same motorcar can be pushed by

densk [106]

Answer:

40N by each person.

Explanation:

8 0

2 years ago

An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa

sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

$T_s = T_e \times \gamma$

where c is the speed of light

v is the speed of the rock star moving

$T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}$

$1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}$

$\sqrt{1-\dfrac{v^2}{c^2}} =0.556$

squaring both side

$1-\dfrac{v^2}{c^2}=0.3086$

$v^2=0.6914c^2$

v = 0.831 c

v = 83.1 % of speed of light

7 0

2 years ago

A race car starting from rest accelerates uniformly at a rate of 3.90 m/s^2 what is the cars speed after it has traveled 200 m.

snow_tiger [21]

The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t = $\sqrt{400/3.9} \approx 10.12739367$ , at = v, so $3.9 * 10.12739367 \approx 39.5 m/s$

6 0

2 years ago

Read 2 more answers