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Romashka-Z-Leto [24]
3 years ago
11

An object moving with constant acceleration changes its speed from 20 m/s to 60 m/s in 2.0 s. What is the acceleration? (b) How

far did it move in this time?
Physics
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

a)20\ \rm m/s^2

b) 80 m

Explanation:

Given:

  • Initial velocity of the object, u=20 m/s
  • Final velocity of the object, v=60 m/s
  • Time taken, t=2 s

Using Equation of motion we have

a)Let a be the acceleration of the object

v=u+at\\60=20+2a\\a=20\ \rm m/s^2

b) Let d be the distance covered by the object given by

Using equation of motion we have

2ad=v^2-u^2\\2\times20\times d=60^2-20^2\\d=80\ \rm m

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The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r
Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength \lambda =1.4cm =0.014m

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m

So distance D will be 49.34 m

4 0
3 years ago
A laser produces light at 5.32x10-7 m.
Blababa [14]

Answer:

D

Explanation:

speed = frequency x wavelength

speed of light in vacuum is 3.0 x 10^8

wavelength = 5.32 x10 ^-7

3.0 x 10 ^ 8 = 5.32 x 10^-7 x frequency

frequency = 5.63909 x 10^14

round it off = 5.64 x 10^14 Hz

thus the answer is D

hope this helps please mark it

6 0
3 years ago
The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

P₂V₂ = n₂RT₂

⇒53530 J = n₂RT₂

\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

8 0
3 years ago
A piece of plastic is uniformly charged with surface charge density eta1. The plastic is then broken into a large piece with sur
Amiraneli [1.4K]
C is the answer for this problem
3 0
3 years ago
A 40 Watt lightbulb is on for 3 minutes. How many joules of work does it do?
ella [17]

Answer:

Explanation:

Power (p)= 40 watt

time (t)= 3 minutes = 180 seconds

Energy (E)= ?

we know

P = E/t

40 = E / 180

E = 7200 Joules

Therefore it does 7200 joules of work.

hope it helps :)

6 0
3 years ago
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