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2 years ago

Can a body having larger mass be hotter than another of smaller mass.expalain

Physics

1 answer:

Rom4ik [11]2 years ago

8 0

Answer:

it would make sense because a larger body could produce more body heat.

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What are some examples of analog and digital signals that we use in our everyday lives?

olga nikolaevna [1]

Answer:

digital signals one non-obvious example of a digital signal is a hard disk. Discrete time signals are defined for a discrete set of values, like the samples of a continuous time signal, while digital signals are inherently digital, like the number that crosses a certain bridge every hour.

Explanation:

Analog signals for example, and aneroid barometer uses rotary position as the signal to conveyed by an analog signal, often such a signal is a measured response to changes in physical phenomena, such as sound, light, temperature, position, or pressure.

5 0

2 years ago

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the o

babymother [125]

(v₂-v₁)/t = a

(v₂-14)/20 = 5

v₂-14 = 100

v₂ = 114 m/s

5 0

2 years ago

Read 2 more answers

Using Newton's 2nd law,find the mass if the force is 18 N and the acceleration is 2 m/s/s.

stich3 [128]

Answer:

9 kg

Explanation:

According to Newton's second law:

$f = ma$

$m = \frac{f}{a}$

Given:

f = 18 N

a = 2 m/s^2

Substitute the given values in the formula.

$m = \frac{18}{2}$

$m = 9$

7 0

2 years ago

A tennis ball is shot straight up from the ground. After traveling 4.00 m, the ball passes a 4.00 m high window and then continu

Sophie [7]

Answer:

a) 24.918 m

b) 4.5078 seconds

c) [0.18, 0.39]

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{20-\frac{1}{2}\times 9.81\times 1^2}{1}\\\Rightarrow u=15.095\ m/s$

At the beginning of the last 1 second the velocity will be 15.095 m/s

Taking that as the final velocity of the path from the maximum height we find time

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{15.095-0}{9.81}\\\Rightarrow t=1.539\ s$

So, the ball traveled for 1+1.539 = 2.2539 seconds from the maximum height to the ground.

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 2.2539^2\\\Rightarrow s=24.918\ m$

The maximum height the ball traveled is 24.918 m.

The time the ball will take to go up is 2.2539 seconds

So, total time the ball will be in flight is 2.2539+2.2539 = 4.5078 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{24.918-\frac{1}{2}\times -9.81\times 2.2539^2}{2.2539}\\\Rightarrow u=22.11\ m/s$