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3 years ago

What is the net Force needed to get a 16 kg box moving 4 m/s^2?

Physics

1 answer:

igomit [66]3 years ago

3 0

Answer:

The net force should be of a magnitude of 64 N

Explanation:

We use Newton's second Law for this:

$F_{net} = m\,*\,a$

which for our case gives:

$F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N$

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An 80.0-kg man jumps from a height of 2.50 m onto a platform mounted on springs. As the springs compress, he pushes the platform

frosja888 [35]

Answer:

6.16 m/s

0.0105 m

Explanation:

Let the ground 0 for potential reference be at where the spring is compress 0.24 m. The the man would jump from a height h = 2.5 + 0.24 = 2.74 m from it. We can apply the law of energy conservation knowing that as the man jumps, his potential energy converts to kinetic energy, then finally to elastic energy:

$E_p = E_e$

$mgh = kx^2/2$

where m = 80 kg is the man mass, g = 9.81 m/s2 is the gravitational acceleration, h = 2.74 m is the potential distance he travels, k N/m is the spring constant and x = 0.24 is the distance it compresses

$80*9.81*2.74 = k0.24^2/2$

$2150.352 = 0.0288k$

$k = 74665 N/m$

Similarly at the position where it compresses by 0.12 m, it's 0.24 - 0.12 = 0.12 m far from ground 0.

$E_p = E_{e2} + E_k + E_{p2}$

$mgh = kx_2^2 + mv^2/2 + mgh_2$

$2150.352 = 74665*0.12^2/2 + 80v^2/2 + 80*9.81*0.12$

$2150.352 = 537.588 + 40v^2 + 94.176$

$40v^2 = 1518.588$

$v^2 = 37.9647$

$v = \sqrt{37.9647} = 6.16 m/s$

When he steps gently, then his gravity force would equal to his spring force

$mg = kx_3$

$x_3 = mg/k = 80*9.81/74665 = 0.0105m$

7 0

3 years ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in va

WINSTONCH [101]

Answer:

Part A: 7500 V

Part B: 2.899×10⁻³ m²

Part C: 10.27 pF or 10.27×10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d................ Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed............. Equation 2

Given: E = 3.0×10⁶ V/m, d = 2.5 mm = 2.5×10⁻³ m

Substitute into equation 2

V = 3.0×10⁶ (2.5×10⁻³ )

V = 7.5×10³ V

V = 7500 V

Part B:

Using,

E = Q/(e₀A).................... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q/(e₀E).............. Equation 4

Given: Q = 77 nC = 77×10⁻⁹ C, E = 3.0×10⁶ V/m

Constant: e₀ = 8.854×10⁻¹² F/m

Substitute into equation 4

A = 77×10⁻⁹/(8.854×10⁻¹²× 3.0×10⁶)

A = 77×10⁻⁹/(26.562×10⁻⁶)

A = 2.899×10⁻³ m²

A = 2.899×10⁻³ m².

Part C:

Using,

Q = CV.................. Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V.............. Equation 6

Given: Q = 77 nC = 77×10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77×10⁻⁹/7500

C = 10.27×10⁻¹² F

C = 10.27 pF

5 0

3 years ago

Jupiter is 317 times more massive than the Earth. An astronaut on Jupiter would _____. weigh less than on Earth weigh more than

Softa [21]

Since "you will weigh more on jupiter" isn't an option, your answer is <u>B."weigh more than on Earth"</u> <em>btw ik this is a late answer but for the people out there still searching for an answer hope this helps!</em>

8 0

3 years ago

A bicyclist steadily speeds up from rest to 11.0m/s during a 5.50s time interval. Determine all unknowns and answer the followin

Marina86 [1]

Explanation:

acceleration is 2 m/s^2

v-u/t

distance travelled is 30.25 meter

(v^2-u^2)/2a

11*11/2*2

121/4

30.25 m

5 0

3 years ago

When you pick a book up off the floor you are tranfering __________ energy from your hands to the ____________ energy of the boo

MakcuM [25]

When you pick a book off the floor you are transferring kinetic energy from your hands to the potential energy of the book.

3 0

4 years ago