All categories

3 years ago

Max ran 100 feet in 10 seconds.

Physics

2 answers:

Ivanshal [37]3 years ago

8 0

Answer:

for max :

100 feet in 10 secs

for molly :

60 feet in 5 secs = 120 feet in 10 secs

so, molly ran farther in the same time interval i.e. covered 120 feet where as Max covered 100 feet

Explanation:

brainliest plz

Pani-rosa [81]3 years ago

5 0

Answer:

1 no Max

2 no Molly

Explanation:

because if molly can run 60 feet in 5 second he can run 120 feet in 10 second

You might be interested in

Describe how freezing could be used to remove sugar from a mixture of sugar and water?

VARVARA [1.3K]

When the mixture (the sugar and water) is frozen, it separates. The water molecules get closer together, separating and pushing the sugar crystals to the top.<span />

7 0

3 years ago

Read 2 more answers

If an engine does 660 J of work in 10 seconds, its average power is ...

givi [52]

Answer:

66w

Explanation:

p=w/t

p=660/10

p=66

prolly a bad explanation but hope it helps...

4 0

3 years ago

A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. How would you describe t

OleMash [197]

Scenes the chair wheels are up the person is rolling backwards and if the wheels were down then the person would go forwards
<span />

3 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

4 years ago

An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use

shtirl [24]

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object = 200 feet/second

Final velocity of object = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

$a=\frac{v_f-v_i}{t}$

where vf represents final velocity, vi represents initial velocity and is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a= -30m/s$

The acceleration of the object is -30m/s

3 0

3 years ago