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3 years ago

If the mass of an object is 8 kg and its momentum is -80 kgm/s, what is its velocity?

Physics

1 answer:

Dimas [21]3 years ago

4 0

An object's momentum is the product of its mass and its velocity:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

p = -80kg×m/s

m = 8kg

Plug in these values and solve for v:

-80 = 8v

v = -10m/s

Choice D

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A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.

Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

$v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s$

Hence, the time interval of the marble is 6.09 seconds.

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3 years ago

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

3 years ago

write a paragraph about convection make sure to include these words -density,increasing,decreasing,rise,sink.

k0ka [10]

Answer:yQIEFUEQf

Explanation:fdgagsdg

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3 years ago

How much time does it take for a rattlesnake to strike its victim if it is strikes at a speed of 1.25 m/s and the victim is exac

Marianna [84]

Answer:

4 seconds

Explanation:

5/1.25=4

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2 years ago

A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d

Minchanka [31]

This problems a perfect application for this acceleration formula:

   Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:   1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by 0.65 m/s²:

   T² = (1600 m) / (0.65 m/s²)

   T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

   T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)

7 0

3 years ago