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2 years ago

1. The head of a rattlesnake can accelerate 50.00 m/s2 in striking a victim. If a car

Physics

1 answer:

scoray [572]2 years ago

3 0

Answer:

2.000 s

Explanation:

Given:

a = 50.00 m/s²

v = 100.0 m/s

v₀ = 0 m/s

Find: t

v = at + v₀

(100.0 m/s) = (50.00 m/s²) t + (0 m/s)

t = 2.000 s

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A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

Strike441 [17]

Answer:

This question can be answered by using conversation of energy.

$K_1 + U_1 = K_2 + U_2$

$\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2$

$\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s$

Explanation:

Note that we take $K_2 = 0$ because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

7 0

3 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago

Calculate the minimum amount of work needed to move a 500-kg rocket payload from Earth's surface to the ISS in orbit 400,000m ab

WINSTONCH [101]

Answer:

2.000.000.000

Explanation:

Apply the formula:

Work = Force . Displacement

W = 500.10 . 400.000 (the 10 come from gravity)

W = 5000 . 400.000

W = 2.000.000.000 Joules

I think it is that, can be wrong.

7 0

2 years ago

A pendulum clock gives correct time at 20°c how many sec/day will it gain or loose when the temperature fall to 5°c? cofficient

rewona [7]

Answer:

129.6 seconds

Explanation:

Given that :

α = 0.0002°c-1

θ1 = 20°C

θ2 = 5°C

Time t = one day ; Converting to seconds ; number of seconds in a day ; (24 * 60 * 60) = 86400 seconds

Let dT= change in time

Using the relation :

dT = 0.5* α * dθ * t

dθ = (20 - 5) = 15°C

dT = 0.5 * 0.0002 * 15 * 86400

dT = 129.6 seconds

5 0

2 years ago