All categories

3 years ago

Which is an example of negative acceleration?

Physics

2 answers:

pshichka [43]3 years ago

7 0

Answer:

A) a snowboarder who increases speed when traveling downhill

just olya [345]3 years ago

4 0

It’s A!!!!!!!!!!!!!!!!!!!

You might be interested in

A man pushes on piano with mass 170 kg; it slides at constant velocity down a ramp that is inclined at 20.0 ∘ above the horizont

nikdorinn [45]

Answer

given,

mass of the piano = 170 kg

angle of the inclination = 20°

moves with constant velocity hence acceleration = 0 m/s²

neglecting friction

so, force required to pull the piano

F = m g sin θ

F = 170 × 9.81 × sin 20°

F = 570.39 N

so, force required by the man to push the piano is F = 570.39 N

4 0

2 years ago

A car accelerates from zero to a speed of 110

Verizon [17]

The car's rate of acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0

3 years ago

Alex787 [66]

It is the subtance betwen the person or the object

7 0

2 years ago

Read 2 more answers

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

2 years ago

A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}$

$d=\frac{1485*5.63}{2}\\d= 4180.3m$

wavelengt of sound is $\lambda$ = v/f

= (1485)/(5920)

= 0.251 m

7 0

3 years ago