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3 years ago

Work out the area of a rectangle with base, b=20mm and perimeter, p=50mm

Physics

2 answers:

Korvikt [17]3 years ago

7 0

Answer:

100mm²

Explanation:

the other side: (50 - 20 × 2) ÷ 2 = 5

5 × 20 = 100

Arada [10]3 years ago

5 0

Answer:

Subtract given lengths: 50 - 20 = 30

Divide by 2 since there are 2 missing sides that need a length: 30 divided by 2 = 15

With the given side lengths multiply all of them to get your area.

15 x 20 = 300 is your area I am pretty sure.

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The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin wi

svlad2 [7]

Answer

given,

largest diameter of balls = 9.52 mm = 0.00476 m

radius = 0.00476

smallest diameter of ball = 2.38 mm = 0.00238 m

radius = 0.00119

viscosity = 1.5 Pa.s

density of the ball = 1.42 g/cm

$F = 6\pi \eta r V_t$

$F = \dfrac{mv}{t}$

$F = \dfrac{\dfrac{4}{3}\pi\ r^3\times (\rho-\sigma) \times 0.99 V_T}{t}$

$6\pi \eta r V_t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{t}$

$t = \dfrac{\dfrac{4}{3}\pi\ r^3\times rho \times 0.99 V_T}{6\pi \eta r V_t}$

$t= \dfrac{0.22 r^2 (\rho-\sigma) }{\eta}$

for small balls

$t= \dfrac{0.22\times 0.00119^2 (1460-1300)}{1.5}$

t = 0.033 ms

for larger ball

$t= \dfrac{0.22\times 0.00476^2 (1460-1300)}{1.5}$

t = 0.531 ms

6 0

4 years ago

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is

Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

$N=mg+\dfrac{mv^2}{r}$

$N=m(g+\dfrac{v^2}{r})$

$N=60\times (9.81+\dfrac{(10)^2}{30})$

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0

3 years ago

If a girl carries 10kg bag

lutik1710 [3]

Then the girl is very strong. 10kg bags are really heavy

6 0

3 years ago

1.The putt shot is used for hitting the golf ball off of the golf tee.

MArishka [77]

Answer:

1.True 2.

Explanation:

3 0

3 years ago

Read 2 more answers

Early one October, you go to a pumpkin patch to select your Halloween pumpkin. You lift the 2.9 kg pumpkin to a height of 1.5 m

Katen [24]

Answer:

Work done in lifting to a height of 1.5 m is 42.63 J.

Work done in carrying it to 50 m is 0 J

Explanation:

Given:

Mass of the pumpkin (m) = 2.9 kg

Vertical displacement of the pumpkin (y) = 1.5 m

Horizontal displacement of the pumpkin (x) = 50 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done by a force is given by the formula:

$W=FS\cos\theta$

Where,

$F\to force\ applied\\\\S\to displacement\ caused\\\\\theta\to angle\ between\ F\ and\ S$

Now, the work done is maximum when both force and displacement is in same direction. ( $\theta=0^\circ$ )
Work done is minimum when both force and displacement are in opposite direction. ( $\theta=180^\circ$ )
Work done is zero when force and displacement are perpendicular to each other. ( $\theta=90^\circ$ )

Here, as the pumpkin is raised to a height 'h', there is force applied against gravity in the upward direction. So, force and displacement are in same direction and thus the angle is 0° between the force and displacement vectors.

So, work done is given as:

$W=Force\times Vertical\ displacement$

Force applied is equal to the gravitational force applied by the Earth but in the opposite direction.

So, Force = mg = $2.9\times 9.8 = 28.42\ N$

So, work, $W=Fy=28.42\times 1.5=42.63\ J$

So, work done in lifting the pumpkin is 42.63 J.

Now, when the pumpkin is carried to the check-out stand, the displacement is horizontal while the force applied is still in the upward direction.

So, the force and displacement are perpendicular to each other. Thus, the work done is zero.

4 0

4 years ago