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3 years ago

As a bicycle is ridden west in a straight line with decreasing speed,the acceleration of the bicycle must be

Physics

1 answer:

jeka943 years ago

3 0

Answer:

Decreasing

Hope this helps! :)

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After travelling 4.0 km, the train reaches its maximum speed. it continues at this constant speed on the next section of the tra

lozanna [386]

Answer:

toward the center of the circular curve

Explanation:

An object will follow a circular path at constant speed if and only if its net acceleration is constant and directed toward the center of the curved path.

The result force is directed toward the center of the circle.

_____

Additional comment

If any part of the result force is in the direction of motion, the speed will not be constant. If the center-directed force is not constant, the path will not be circular.

5 0

2 years ago

An acorn falls from a tree. Its velocity just before it hits the ground is 28.2 m/s, downward. (acceleration of gravity is 9.81m

ss7ja [257]

Answer:

12.74 ms^-1 download

Explanation:

v=28.2, a=9.81

start from rest u=0

v=u+at=0+(9.81)t=28.2

t=2.875...

it reach 1.4 second before hitting the ground:

t=1.4, u=0, a=9.81

v=u+at=0+(9.81)(1.4)=12.74

7 0

3 years ago

N 1800kg car has an of 3.8m/s? What is it on the car? acceleration force acting

nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

$a=3.8 m/s^2$

Substituting,

$F=(1800)(3.8)=6840 N$

7 0

3 years ago

The refractive index of glass is 1.52

Ksenya-84 [330]

Explanation:

..upper side is glass ..

3 0

3 years ago

Astronaut Sarah leaves Earth in a spaceship at a speed of 0.280c relative to Earth. Sarah's destination is a star-system 12.5 li

Usimov [2.4K]

Answer:

L= 12 light years

Explanation:

for length dilation we use the formula

$L=L_0\sqrt{1-\frac{v^2}{c^2} }$

now calculating Lo

Lo = 12.5×365×24×3600×3×10^8

= 1.183×10^17 m

now putting the values of v and Lo in the above equation we get

$L=1.183\times10^{17}\sqrt{1-\frac{0.28c^2}{c^2} }$

= 1.136×10^17 m

L= $= \frac{1.136\times10^{17}}{365\times24\times3600\times3\times10^8}$ m

so L= 12 light years

8 0

4 years ago