A car started from a rest and accelerated at 9.54 m/s2 for 6.5 seconds. How much distance was covered by the car?

After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume t

Ivenika [448]

Time taken by the package to reach the sea level= 13.7 s

height=h=925 m

initial velocity along vertical= vi=0

acceleration due to gravity=g=9.8 m/s^2

using the kinematic equation h= Vi*t + 1/2 gt^2

925=0(t)+1/2 (9.8)t^2

4.9 t^2=925

t= 13.7 s

6 0

3 years ago

A transverse wave on a string is described by the wave function

svp [43]

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

Read more on transverse wave here

brainly.com/question/2516098

#SPJ4

5 0

1 year ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

3 0

3 years ago

A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

$p=\frac{F}{A}$

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

$p\propto \frac{1}{A}$

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0

3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago