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Nimfa-mama [501]
3 years ago
11

A television weighs 8.50 pounds. How many grams is this? (Hint: You need to

Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

3859 grams

Explanation:

Given: Weight of a television = 8.50 pounds

To find: Weight of a television in grams

Solution:

1 pound = 0.454 kg and 1 kg = 1000 g

So,

1 pound = 0.454 × 1000 = 454 grams

8.50 pounds = 8.50 × 454 = 3859 grams

Therefore,

Weight of television in grams = 3859 grams

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Explanation:

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What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl
gtnhenbr [62]
An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.
5 0
4 years ago
A 2kg rock is moving at a speed of 6m/s. What constant force is needed to stop the rock in 7 x 10^-4?
dsp73

Explanation:

key to this problem is the impulse-momentum theorem which states that the change in the momentum of an object is equal to the impulse applied into it.

J

=

Δ

p

,

where

J

is the impulse and

Δ

p

is the change in momentum. Basically, the impulse is the product of force and time duration, that is,

J

=

F

Δ

t

In this problem, the impulse would be the product of the force stopping the rock and

0.7

s

.

On the other hand, momentum

p

is the product of the mass

m

and velocity

v

. Therefore, the change in momentum is given by

Δ

p

=

m

2

v

2

−

m

1

v

1

.

Starting with the impulse-momentum equation, we have

J

=

Δ

p

F

Δ

t

=

m

2

v

2

−

m

1

v

1

Divide both sides by

Δ

t

,

we get

F

Δ

t

Δ

t

=

m

2

v

2

−

m

1

v

1

Δ

t

F

=

m

2

v

2

−

m

1

v

1

Δ

t

Finally, substitute the values and we get

F

=

(

2

kg

)

(

0

)

−

(

2

kg

)

(

6

m

s

)

(

0.7

s

)

F

≈

−

20

kg

m

s

2

Since

1

N

=

1

kg

m

s

2

,

then

F

≈

−

20

N

Therefore, using the correct significant figures (in this case, we need one significant figure since 2 kg, 6 m/s and 0.7 s all have one) in the final answer, we would need to have approximately

20

N

force to stop the rock in

0.7

s

.

Note: The negative sign is referring to the direction of the force opposite of the direction of the velocity

v

1

.

6 0
3 years ago
A bag is dropped from a balloon that is 50m above the ground and rising at 15m/s. calculate
ANTONII [103]

Answer:

See the answers below

Explanation:

We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.

For this case we have mechanical energy, which is the sum of the kinetic and potential energies.

E_{1}=E_{2}

where:

E_{1}=E_{k}+E_{p}\\E_{2}=E_{p}

Ek = kinetic energy [J] (units of Joules)

Ep = potential energy [J]

In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.

A)

m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]

B)

With the value calculated above we can find the acceleration of the balloon.

The distance traveled is the difference between the maximum height and 50 meters.

x = 61.46-50\\x = 11.46[m]

With the following equation of kinematics.

v_{f}^{2} =v_{o}^{2}+2*a*x

0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]

The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.

We can use the following equation of kinematics to find the final velocity after 4 seconds.

v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]

Now the distance:

v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]

c) Using the following equation of kinematics.

v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]

4 0
3 years ago
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