Ask question

Ask question

All categories

3 years ago

7

A 150-kg object takes 1.5 minutes to travel a 2,500- meter straight path. It begins the trip traveling 120 m/s and decelerates t

o a velocity of 20 m/s . What is the acceleration?

1 answer:

lana [24]3 years ago

4 0

The answer is it is -1.11 m/s^2.

You might be interested in

a merry go round has a radius of 8 meters makes 2 revolutions every 2.5 minutes. A. express the angular speed of the merry go ar

pantera1 [17]

Answer:

a). $v=83.77x10^{-3} rad/s$

b). $v=0.8$ rpm

c).v=0.5865 m/sec.

Explanation:

Given:

$r=8m$

$v=\frac{2rev}{2.5minutes}$

a).

$2 rev*\frac{2\pirad}{1rev}=4\pi rad$

$t=2.5minutes*\frac{60s}{1minute}=150s$

The angular speed in radians per seconds is

$v=\frac{4\pi}{150s}=83.77x10^{-3} rad/s$

b).

$v=\frac{2rev}{2.5minute}rpm$

$v=0.8$ rpm

c)

Child's distance per revolution

(pi*2r) = 43.988 metres.

v=(43.988 x 0.0133333) = 0.5865 m/sec.

4 0

3 years ago

If you throw a ball up with a velocity of 7 m/s, how long will it take for the ball to reach

AveGali [126]

Answer:

42secs

Explanation:

v=u-gt

0=7-10t

10t=7

t=7/10

t=7/10*60

t=42

7 0

2 years ago

A net force of 16 N causes a mass to accelerate at a rate of 5 m/s^2. Determine the mass.

Gnom [1K]

Answer:

So mass of the object will be 3.2 kg

Explanation:

We have net force on the object F = 16 N

Acceleration of the object $a=5m/sec^2$

We have top find the mass of the object

From newton law of motion we know that force is given by

F = ma , here m is mass and a is acceleration

So $16=5\times m$

$m=3.2kg$

So mass of the object will be 3.2 kg

3 0

3 years ago

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

3 years ago

When the k. E of

Ierofanga [76]

$\sqrt{2}$ Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = $\sqrt{2}$ V1

Since momentum = M V the momentum increases by $\sqrt{2}$

8 0

3 years ago

Read 2 more answers