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svlad2 [7]
3 years ago
14

VA ROG AJUTATIMA ,VA DAU ORICE ,VA ROG AJUTATIMA

Physics
1 answer:
Brums [2.3K]3 years ago
4 0

I agree.......................

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Can someone please help
ki77a [65]

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

8 0
3 years ago
A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0
4 years ago
An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.
Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

4 0
3 years ago
Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon
Mice21 [21]

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

6 0
3 years ago
Explain how a rotating electric fan might be used to model the atom. explain how the rotating fan is unlike an atom.
quester [9]

A fan may be used to model an atom because the fan blades form something similar to a electron cloud. But there is really nothing to model the neutrons and protons. Also the fan is flat like where an atom is more 3-D sphere-like.



hope this helps


Ur welcome


have a great day (:

3 0
4 years ago
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