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sleet_krkn [62]

2 years ago

6

A girl is running toward the front of a train at 10 m/s. If the train is going 75 m/s on the Southbound tracks, what is the spee

d of the train relative to the girl?

2 answers:

sineoko [7]2 years ago

6 0

Answer:

velocity of the train with respect to girl is 85 m/s

Explanation:

As we know that girl is moving towards the train in front of it with speed

$v_1 = 10 m/s$

also we know that the speed of the train towards the girl is

$v_2 = 75 m/s$

Now we know that relative velocity of the train with respect to girl is given as

$\vec v_{tg} = \vec v_t - \vec v_g$

here let say the velocity of train is positive while girl is running opposite to train so its velocity is negative

$\vec v_{tg} = 75 m/s - (-10 m/s)$

$\vec v_{tg} = 85 m/s$

dimulka [17.4K]2 years ago

5 0

We can take as positive direction the direction of the train, and as negative direction the direction of the girl. The speed of the train relative to the girl will be given by the difference between the two velocities:
$v' = v_t - v_g$
which means
$v'=75 m/s - (-10 m/s)=85 m/s$

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Oksanka [162]

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7 0

2 years ago

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they? (

valkas [14]

The gravitational force between two masses m₁ and m₂ is
$F=G \frac{m_{1} m_{2}}{d^{2}}$
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d = distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
= 0.0222
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Answer: 149.2 mm

8 0

2 years ago

If current I=2A and resistance R= 5 ohms what is the potential difference (voltage) V?

damaskus [11]

Answer:

10 V

Explanation:

From Ohm's law, V=IR where V is the potential difference voltage, I is the current and R is resistance. Substituting 2A for current and 5 Ohms for resistance then V=2*5= 10 V

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2 years ago

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned

liq [111]

Answer:

$T = 3183 N$

Explanation:

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$\Delta L = 2(2\pi r)$

$\Delta L = 4\pi r$

$\Delta L = 4(\pi)(1.0 mm)$

$\Delta L = 12.56 mm$

now we know by the formula of Young's modulus

$Y = \frac{T/A}{\Delta L/L}$

so we have

$T = \frac{AY \Delta L}{L}$

$T = \frac{\pi (0.55 \times 10^{-3})^2(2.0 \times 10^{11})(12.56 \times 10^{-3})}{0.75}$

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3 0

2 years ago

When a drag strip vehicle reaches a velocity of 60 m/s, it begins a negative acceleration by releasing a drag chute and applying

hammer [34]

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ax$ (1)

Where:

x: is the displacement

$v_{f}$ : is the final speed = 0 (reduces its velocity back to zero)

$v_{0}$ : is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

$x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m$

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

5 0

2 years ago