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blondinia [14]
3 years ago
8

Fgoods3eddddddddddddddd

Physics
2 answers:
KonstantinChe [14]3 years ago
7 0

Answer: beans greens potatoes tomatoes lambs hams you name it

Explanation:

dem82 [27]3 years ago
6 0

Answer:

ejidhddjjfbrfofjfbjf

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Spiders kan swim???????
Nastasia [14]

Answer:

Spiders cannot actually propel their bodies through the water as a swimmer does, but they can use objects to get across the water and some can run across the water.

Explanation:

3 0
3 years ago
Read 2 more answers
a missile is moving 1810 m/s at a 20.0 degree angle. it needs to hit a target 19,500 m away in a 32.0 degree direction in 9.20 s
Ket [755]

Answer:

112 m/s², 79.1°

Explanation:

In the x direction, given:

x₀ = 0 m

x = 19,500 cos 32.0° m

v₀ = 1810 cos 20.0° m/s

t = 9.20 s

Find: a

x = x₀ + v₀ t + ½ at²

19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²

a = 21.01 m/s²

In the y direction, given:

y₀ = 0 m

y = 19,500 sin 32.0° m

v₀ = 1810 sin 20.0° m/s

t = 9.20 s

Find: a

y = y₀ + v₀ t + ½ at²

19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²

a = 109.6 m/s²

The magnitude of the acceleration is:

a² = ax² + ay²

a² = (21.01)² + (109.6)²

a = 112 m/s²

And the direction is:

θ = atan(ay / ax)

θ = atan(109.6 / 21.01)

θ = 79.1°

5 0
3 years ago
A pursuit spacecraft from the planet Tatooine is attempting to catch up with a Trade Federation cruiser. As measured by an obser
Ede4ka [16]

Answer:

0.384c

Explanation:

To find the speed of the pursuit ship relative to the cruiser you use the following relativistic equation:

u'=\frac{u-v}{1-\frac{uv}{c^2}}

u': relative speed

u: speed of the pursuit ship = 0.8c

v: speed of the cruiser = 0.6c

c: speed of light

You replace the values of the parameters to obtain u':

u'=\frac{0.8c-0.6c}{1-\frac{(0.6c)(0.8c)}{c^2}}=0.384c

Hence, the relative speed is 0.384c

4 0
3 years ago
What do period and frequency both describe?
Natasha2012 [34]
Frequency refers to how often something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity. Period is a time quantity. Frequency is the cycles/second
8 0
3 years ago
If the car’s speed decreases at a constant rate from 60 mi/hmi/h to 50 mi/hmi/h in 3.0 ss, what is the magnitude of its accelera
Kitty [74]

Answer:

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

Explanation:

Hi there!

The acceleration is defined as the change in velocity in a time:

a = Δv / Δt

Where:

a = acceleration.

Δv = change in velocity  = final velocity - initial velocity.

Δt = elapsed time.

In this case:

Initial velocity = 60 mi/h

final velocity = 50 mi/h

elapsed time = 3.0 s

Let´s convert the time unit into h:

3.0 s · 1 h /3600 s = 1/1200 h

Now, let´s calculate the acceleration:

a = Δv / Δt

a = (50 mi/h - 60 mi/h) / 1/1200 h

a = -1.2 × 10⁴ mi/h²

The magnitude of the acceleration is 1.2 × 10⁴ mi/h²

7 0
3 years ago
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