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3 years ago

Three-dimensional ____ programs allow designers to rotate designs of 3-d objects to view them from any angle.

Physics

1 answer:

notsponge [240]3 years ago

5 0

Long phrase is the blank

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11. You want to calculate the displacement of an object thrown over a bridge. Using -10 m/s2 for acceleration due to gravity, wh

IrinaVladis [17]

Answer:

The displacement was 320 meters.

Explanation:

Assuming projectile motion and zero initial speed (i.e., the object was dropped, not thrown down), you can calculate the displacement using the kinematic equation:

$d = \frac{1}{2}gt^2=\frac{1}{2}10\frac{m}{s^2}\cdot 8^2 s^2=320 m$

The displacement was 320 meters.

7 0

3 years ago

A train travels 160 km in 2 h. What is the train’s average speed in km/h?

loris [4]

The train’s average speed is 80km/h

3 0

3 years ago

Read 2 more answers

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

Could the half of the moon that faces the earth ever be completely dark in any of these diagrams

docker41 [41]

Yes well maybe but I think yes

6 0

3 years ago

The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by

Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

$m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}$

we have

a. for c = 20 cm

$m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,$

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

$\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.$

Evaluated at c = 20 cm, we have

$\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.$

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0

3 years ago