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MaRussiya [10]
3 years ago
12

A 1000 kg car pushes a 2000 kg car that has a dead battery. When the driver steps on the accelerator, the drive wheels of the ca

r push backward against the ground with a force of 4500 N.
Physics
1 answer:
butalik [34]3 years ago
4 0
So what is the question
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A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?
Irina-Kira [14]
F = 400 Pa x 2.5 m2
F = 1 kN
4 0
2 years ago
2.5 gram sample of a radioactive element was formed in a 1960 explosion of an atomic bomb at Johnson Island in the Pacific Test
kow [346]

Answer:

0.15625 grams

Explanation:

Half life: It is related to the decay of radioactive material. The duration in which  half of the material will be degraded/decayed. That means after half life 50% of the radioactive material will be left. Here the half life is 28 years.

Initial quantity of the sample: 2.5 grams.

After 28 years, the leftover quantity = 1.25 grams

After 56 years, the leftover quantity = 0.625 grams

After 84 Years, the leftover quantity = 0.3125 grams

After 112 years, the leftover quantity = 0.15625 grams

5 0
3 years ago
The increase in height as a tsunami approaches shore is due to
ryzh [129]

Answer:

C. The decrease in speed as the wave approaches shore.

Explanation:

The waves break when approaching the shore because the depth decreases. Thus, the wave travels more slowly and increases its height. There comes a time when the part of the wave on the surface travels faster than the one that travels under water, the ridge destabilizes and falls against the ground.

5 0
3 years ago
Read 2 more answers
An object of mass 25kg is falling from the height h=10 m. calculate
r-ruslan [8.4K]

Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

V²=U²+2gh

V²=0 + 2×10×10=200m/s

a).kinetic energy=(1/2)mv²=(1/2)25×200=2500

potential energy=mgh

p.e=25×10×10=2500J

pe+ke=2500+2500=5KJ

b).mgh=25×10×4=1000J

c). V²=U²+2gh

V²=0+2×10×4

V²=80

kinetic energy=(1/2)mv²

=(1/2)25×80

=1KJ

d). From my first paragraph V²=200

V=√200

V=14.142m/s

6 0
3 years ago
A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If
Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0
3 years ago
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