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balu736 [363]
2 years ago
13

21. The force is the product of

Physics
1 answer:
KiRa [710]2 years ago
4 0

Answer:

D mass and acceleration

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A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby has a mass of 4kg. The carriage ha
Ira Lisetskai [31]

Answer:

E=252J

Explanation:

The total mechanical energy of an object or system is given by:

E mech=K+U

Where K is the kinetic energy of the object and U is the potential energy of the object. The carriage, sitting motionless at the top of the hill, has only potential energy in the form of gravitational potential energy.

Gravitational potential energy is given by:

Ug=mgh

Where m is the mass of the object, g is the gravitational acceleration constant, and h is the height of the object above some specific reference point, in this case the ground 21 m below.

The weight of a stationary object at the surface of the earth is equal to the force of gravity acting on the object.

W=→Fg=mg

We are given that the carriage weighs 12 N, therefore mg=12N.

Ug=12N⋅21m

⇒Ug=252Nm=252J

Hope it helped, God bless you!

5 0
3 years ago
What is the difference between celestial and terrestrial in planetary terms? Someone please answer
andriy [413]

"Celestial"  =  anything to do with the sky

("Cielo" ..... Spanish for "sky"
 "Ceiling" ... that thing up over your head
 "Caelum" .. Latin for "heaven")


"Terrestrial"  =  anything to do with the Earth

("Terra" ... Latin for "Earth")
5 0
3 years ago
What magnitude charge creates a 1.0 n/c electric field at a point 1.0 m away?
Stolb23 [73]

Answer:

1.1\cdot 10^{-10}C

Explanation:

The electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have

E = 1.0 N/C (magnitude of the electric field)

r = 1.0 m (distance from the charge)

Solving the equation for q, we find the charge:

q=\frac{Er^2}{k}=\frac{(1.0 N/c)(1.0 m)^2}{9\cdot 10^9 Nm^2c^{-2}}=1.1\cdot 10^{-10}C

8 0
3 years ago
A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th
N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0
3 years ago
The eiffel tower is a steel structure whose height increases by 19.5 cm when the temperature changes from −8 to +42 °c. what is
Readme [11.4K]
 The coefficient of expansion is 13 * 10^-6 m per meter length.per oK 
The temperature difference = 42 - - 8 = 50 oC 
delta T = (42 + 273) - (-8 + 273) = 50 oK 
delta L = L * 13* 10^6 m/oK 
oK = 50 oK delta L = 19.5 cm = 19.5 cm [1m / 100 cm] = 0.195m 
So we need to find the length and it is computed by:
0.195= L * 13 * 10^-6 * 50 L = 0.195 / (13*10^-6*50) L = 300 m 
6 0
3 years ago
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