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1 year ago

Please help me! Uniform acceleration problem sheet:

Physics

1 answer:

Brrunno [24]1 year ago

7 0

From the calculation, the value of the acceleration is 5.8 m/s^2.

<h3>What is uniform acceleration?</h3>

The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.

Given the fact that the car started from rest and reached a velocity of 780.34 mph or 348.84 m/s in 1 minute of 60 seconds;

v = u + at

a = v/t

a = 348.84 m/s/ 60 seconds

a = 5.8 m/s^2

Learn more about acceleration:brainly.com/question/12550364?

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When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point

Inessa [10]

Answer:

Explanation:

Given

angle through which ball is launched $=45^{\circ}$

Range of ball=50 m

Range of projectile is $=\frac{u^2sin2\theta }{g}$

$50=\frac{u^2sin90}{9.8}$

u=22.136 m/s

If ball is thrown straight upward

$v^2-u^2=2as$

$0-(22.136)^2=2(-9.8)s$

$s=\frac{22.136^2}{2\times 9.8}$

s=25 m

(b)For Projectile time of flight is

$t=\frac{2usin\theta }{g}$

$t=\frac{2\times 22.136\times sin45}{9.8}$

t=3.19 s

7 0

2 years ago

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5 0

2 years ago

. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio

Reil [10]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.

In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.

The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.

For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.

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2 years ago

An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.

horrorfan [7]

Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

$f = \frac{R}{2}$

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

$f = \frac{24\ cm}{2}$

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2 years ago

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Ganezh [65]

The answer is D that is the right question

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2 years ago