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3 years ago

A baby elephant weighs 250 pounds on Earth. How much would the elephant weigh on Saturn?

1 answer:

8 0

Well since staurns gravity is 10.44 multiply the elephants weight times the gravity
250x10.44 you will get 2610
hope i helped add me
Dosvedanya :)

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A person notices a mild shock if the current along a path through the thumb and index finger exceeds 82 µA. Find the maximum all

Drupady [299]

Answer:

$V_{voltage}=19.68V$

Explanation:

Given data

Current I=82µA=82×10⁻⁶A

Resistance R=2.4×10⁵Ω

to find

Voltage

Solution

From Ohms law we know that:

$V_{voltage}=I_{current}*R_{resistance}\\V_{voltage}=82*10^{-6}*2.4*10^{5} \\V_{voltage}=19.68V$

4 0

3 years ago

According to the Natural Resources Defense Council, what is the largest contributor to land pollution?

Anika [276]

Answer: Food

Explanation:

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2 years ago

Which of the following is an example of a conductor?

nordsb [41]

Answer: B. a gold chain

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2 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

Ahmad is riding his bicycle. He finds that he can accelerate from rest at 0.44 m/s^2 for 5 s to reach a speed of 2.2 m/s. The to

snow_lady [41]

Answer:

1) The force Christian can exert on his bicycle before picking up the the cargo is 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo is 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo

Explanation:

The given parameters are;

The mass of Christian and his bicycle = 54 kg

The mass of the cargo = 12 kg

1) The force Christian can exert on his bicycle before picking up the the cargo = Mass of Christian and his bicycle × Acceleration due to gravity

∴ The force Christian can exert on his bicycle before picking up the the cargo = 54 kg × 9.81 m/s² = 529.74 N

2) The force Christian can exert on his bicycle after picking up the the cargo = (54 + 12) kg × 9.81 m/s² = 647.46 N

Therefore, Christian has to exert more force on his bike after picking up the cargo.

7 0

2 years ago