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3 years ago

Please need help on this not too sure on this

Physics

2 answers:

Bezzdna [24]3 years ago

8 0

The last one because if the lady bug is small it will look big looking through the magnify glass

n200080 [17]3 years ago

3 0

Answer:

the last one

Explanation:

Because it is a magnifying glass, it magnifies the object and makes it bigger than it appears

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What features were the last to form on the moon?

kap26 [50]

<span>Rayed craters-</span><span> were the last features to form on the moon.</span>

3 0

3 years ago

Read 2 more answers

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th

Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0

3 years ago

7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.

Galina-37 [17]

Answer:

a) 0.05s

b) 4000N

Explanation:

a)When car is stopped its final velocity become zero

U- 10 m/s

V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

U = 10

t = 0.05 s

a =?

V = U + at

0 = 10 -a * 0.05

a = 200 m/s2

F = m *a

= 20 * 200

= 4000N

6 0

3 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

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3 years ago

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

Degger [83]

Answer:

147.456077993 Hz

Explanation:

$f_0$ = Frequency of the sonar = 22 kHz