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kiruha [24]
3 years ago
13

What amount of force is needed to propel and object of 27 kg to an acceleration of 11,550 m/s^2? (1 point)

Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer:

311,850 N

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force applied on an object

m is the mass of the object

a is its acceleration

For the object in this problem,

m = 27 kg

a=11550 m/s^2

Substituting, we find the force required:

F=(27)(11550)=311,850 N

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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
3 years ago
HELPPPPPPP PLEASE!!! does anybody have the answer key to edge. physics lab report : mechanical equivalent to heat? one page of d
Karo-lina-s [1.5K]

Answer:

the mechanical equivalent of heat states that motion and heat are mutually interchangeable and that in every case

Explanation:

6 0
3 years ago
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Calculate the speed of a car that has traveled 300 miles in 6 hours.
PilotLPTM [1.2K]
300 miles / 6 hours = 50 miles per hour
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3 years ago
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

To learn more about Attention here:

brainly.com/question/28500124

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6 0
1 year ago
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