Fifty points please help?

A body travels 80 meters in 8 seconds <br> at a steady speed calculate it's average speed<br>

iVinArrow [24]

Explanation:

avg spd = total distance/total time

=80/8=10m/s

7 0

4 years ago

Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side

pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0

3 years ago

Two blocks of masses 3.0 kg and 5.0 kg are connected by a spring and rest on a frictionless surface. They are given velocities t

miskamm [114]

Answer:

-0.7 m/sec

Explanation:

Mass of first block = m1 =3.0 kg

Mass of second block = m2= 5.0 kg

Velocity of first block = V1= 1.2 m/s

Velocity of second block = V2 = ?

Momentum of Center of mass MVcom is sum of both blocks momentum and is given by

MVcom= m1v1+m2v2

Where

M= mass of center of mass

Vcom= Velocity of center of mass=0 m/s (because center of mass is at rest , so Vcom = 0 m.sec)

Putting values, we get;

0= 3×1.2+5v2

==> v2= 3.6/5= - 0.7 m/s

-ve sign indicates that block 2 is moving in opposite direction of block 1

3 0

3 years ago

Which one is it<br> Plz help me :)

Gennadij [26K]

Answer:

5 I think will be none of the above and 6 could be all of the above

8 0

3 years ago

A bucket of mass m is hanging from the free end of a rope whose other end is wrapped around a drum (radius R, mass M) that can r

mr Goodwill [35]

Answer:

Explanation:

Let T be the tension .

Applying newton's second law on the downward movement of the bucket

mg - T = ma

On the drum , a torque of TR will be acting which will create an angular acceleration of α in it . If I be the moment of inertia of the drum

TR = Iα

TR = Ia/ R

T = Ia/ R²

Replacing this value of T in the other equation

mg - T = ma

mg - Ia/ R² = ma

mg = Ia/ R² +ma

a ( I/ R² +m)= mg

a = mg / ( I/ R² +m)

mg - T = ma

mg - ma = T

mg - m x mg / ( I/ R² +m) = T

mg - m²g / ( I/ R² +m ) = T

mg - mg / ( 1 + I / m R² ) = T

b ) T = Ia/ R²

I = TR² / a

c ) Moment of inertia of hollow cylinder

I = 1/2 M ( R² - R² / 4 )

= 3/4 x 1/2 MR²

= 3/8 MR²

I / R² = 3/8 M

a = mg / ( I/ R² +m)

a = mg / ( 3/8 M + m )

T = Ia/ R²

= 3/8 MR² x mg / ( 3/8 M + m ) x 1 /R²

= $\frac{3mMg}{(3M +8m)}$

7 0

4 years ago