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3 years ago

15

A cart was pulled for a distance of 1 kilometer and the amount of work accomplished equaled 40,000 joules. With what force was t

he work accomplished?

1 answer:

goblinko [34]3 years ago

3 0

Answer:

40 N

Explanation:

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VARVARA [1.3K]

Answer:

the correct one is D,

Ultraviolet, x-ray, gamma ray

Explanation:

Electromagnetism radiation are waves of energy that is expressed by the Planck relationship

E = h f

where h is the plank constant and f the frequency of the radiation.

Also the speed of light is

c = λ f

we substitute

E = h c /λ

therefore to damage the cells of the body radiation of appreciable energy is needed

microwave radiation has an energy of 10⁻⁵ eV

infrared radiation E = 10⁻² eV

visible radiation E = 1 to 3 eV

radiation Uv E = 3 to 6 eV

X-ray E = 10 eV

gamma rays E = 10 5 eV

therefore we see that the high energy radiation is gamma rays, x-rays and ultraviolet light.

When checking the answers, the correct one is D

6 0

3 years ago

An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?

Archy [21]

Answer:

4m/s^2

Explanation:

mass(m)=20 kg

force=80 N

acceleration (a)=?

Therefore,

Force = mass * acceleration

80 = 20*a

a=80/20

=4m/s^2

7 0

2 years ago

What if the height is 85m, then how long long does it stay in the air?

wolverine [178]

Explanation:

hamburger cheesecake

6 0

3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

4 years ago

The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele

Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 $m/sec^2$

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion $v^2=u^2+2as$

$553^2=0^2+2\times a\times 0.855$

$a=178835.6m/sec^2$

5 0

4 years ago