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viktelen [127]
3 years ago
5

An ant is crawling along a straight wire, which we shall call the x axis, from A to B to C to D

Physics
1 answer:
Inga [223]3 years ago
5 0
A high quality insecticide applied all along the axis and the surrounding area can protect against a recurrence for a substantial period of time.
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How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

d=vt

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

d=(2.1)(10)=21 m

3 0
3 years ago
What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?
Alisiya [41]
We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
\Delta V = I R
where
\Delta V is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and R=330 \omega, so the potential difference is
\Delta V = IR=(4.00 A)(330 \omega)=1320 V
7 0
3 years ago
A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc
max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }

f = \frac{1}{2\pi } ( 13.522)

f = 2.1535 Hz

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0
1 year ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
Electric charges can only be transferred through a process called conduction.true or false
umka21 [38]

I'm going to say false hope that helped

5 0
3 years ago
Read 2 more answers
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