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AleksandrR [38]
3 years ago
13

A 58 kg boy and a 38 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio

n of the girl toward the boy is 2.4 m/s 2 , determine the magnitude of the acceleration of the boy toward the girl. Answer in units of m/s 2 .
Physics
1 answer:
Arada [10]3 years ago
6 0
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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball
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I need an answer choice
8 0
3 years ago
A natural atom can be negatively charged by
svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

4 0
2 years ago
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A rock climber hangs freely from a nylon rope that is 15 m long and has a diameter of 8.3 mm. If the rope stretches 5.1 cm, what
irinina [24]

Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        \Delta L=\frac{PL}{AE}

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg  

Mass of the climber = 69.38 kg

6 0
3 years ago
You are at a train station, standing next to the train at the front of the first car. The train starts moving with constant acce
Aleksandr [31]

The time needed for the 7th car to pass is 13.2 s

Explanation:

The motion of the train is a uniformly accelerated motion, therefore we can use suvat equations.

We start by analzying the motion of the first car, by using the equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered by the first car in a time t, which corresponds to the length of one car

u = 0 is the initial velocity

a is the acceleration

t = 5.0 s is the time

The equation can be rewritten as

a=\frac{2s}{t^2}=\frac{2L}{(5.0)^2}=0.08L[m/s^2]

where L is the length of one car.

The same equation can be written considering the first 7 cars:

7L = ut+\frac{1}{2}at'^2

where

7L is the distance covered by the 7 cars

t' is the time needed

We still have

u = 0

And the acceleration is constant so it is

a=0.08L

Substituting into the equation, we can find t':

7L = \frac{1}{2}(0.08L)t'^2\\7=0.04t'^2\\t'=\sqrt{\frac{7}{0.04}}=13.2 s

In attachment the graph of the distance covered versus the time taken: since the motion is uniformly accelerated, the relationship between the two variables is quadratical.

Learn more about accelerated motion:

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brainly.com/question/11181826

brainly.com/question/2506873

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8 0
3 years ago
A circuit contains a single 270-pf capacitor hooked across a battery. it is desired to store four times as much energy in a comb
yan [13]
The energy stored by a system of capacitors is given by
U= \frac{1}{2}C_{eq} V^2
where Ceq is the equivalent capacitance of the system, and V is the voltage applied.

In the formula, we can see there is a direct proportionality between U and C. This means that if we want to increase the energy stored by 4 times, we have to increase C by 4 times, if we keep the same voltage.

Calling C_1 = 270 pF the capacitance of the original capacitor, we can solve the problem by asking that, adding a new capacitor with C_x, the new equivalent capacitance of the system C_{eq} must be equal to 4C_1. If we add the new capacitance X in parallel, the equivalent capacitance of the new system is the sum of the two capacitance
C_{eq} = C_1 + C_x
and since Ceq must be equal to 4 C1, we can write
C_1+C_x = 4C_1
from which we find
C_x=3C_1=3 \cdot 270 pF=810 pF
5 0
3 years ago
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