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2 years ago

What is most often given a value of zero to describe an object position on a srtaight line

Physics

1 answer:

Vsevolod [243]2 years ago

6 0

Reference point is most often given a value of zero to describe an object's position on a straight line

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Which of the following items best embodies the physical property of conductivity?

Sladkaya [172]

C is the correct answer

7 0

2 years ago

Read 2 more answers

A spring is hung from the ceiling. A 0.442-kg block is then attached to the free end of the spring. When released from rest, the

siniylev [52]

Answer:

$K=58.8N/m$

Explanation:

From the question we are told that:

Mass $M=0.442$

Drop distance $d=0.150$

Generally the equation for Spring Constant is mathematically given by

$K=\frac{2mg}{x}$

$K=\frac{2*0.442*9.8}{1.150}$

$K=58.8N/m$

6 0

2 years ago

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire

Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$

3 0

2 years ago

A 300 g glass thermometer initially at 23 ◦C is put into 236 cm3 of hot water at 87 ◦C. Find the final temperature of the thermo

DIA [1.3K]

Answer:

$74^{\circ} C$

Explanation:

We are given that

Mass of glass, $m=300 g$

$T_1=23^{\circ}$

Volume, $V=236cm^3$

Mass of water= $density\times volume=1\times 236=236 g$

Density of water= $1g/cm^3$

Temperature of hot water, $T=87^{\circ}$

Specific heat of glass, $C_g=0.2cal/g^{\circ}C$

Specific heat of water, $C_w=1 cal/g^{\circ}C$

$Q_{glass}=m_gC_g(T_f-T_1)=300\times 0.2(T_f-23)$

$Q_{water}=m_wC_w(T_f-T)=236\times 1(T_f-87)$

$Q_{glass}+Q_{water}=0$

$300\times 0.2(T_f-23)+236\times 1(T_f-87)$

$60T_f-1380+236T_f-20532=0$

$296T_f=20532+1380=21912$

$T_f=\frac{21912}{296}=74^{\circ} C$

5 0

2 years ago

A physics student skis down a hill, accelerating at a constant

ikadub [295]

<h3>Answer:</h3>

225 meters

<h3>Explanation:</h3>

Acceleration is the rate of change in velocity of an object in motion.

In our case we are given;

Acceleration, a = 2.0 m/s²

Time, t = 15 s

We are required to find the length of the slope;

Assuming the student started at rest, then the initial velocity, V₀ is Zero.

<h3>Step 1: Calculate the final velocity, Vf</h3>

Using the equation of linear motion;

Vf = V₀ + at

Therefore;

Vf = 0 + (2 × 15)

= 30 m/s

Thus, the final velocity of the student is 30 m/s

<h3>Step 2: Calculate the length (displacement) of the slope </h3>

Using the other equation of linear motion;

S = 0.5 at + V₀t

We can calculate the length, S of the slope

That is;

S = (0.5 × 2 × 15² ) - (0 × 15)

= 225 m

Therefore, the length of the slope is 225 m

6 0

2 years ago