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love history [14]
3 years ago
15

A 1.35 kg block at rest on a tabletop is attached to a horizontal spring having constant 19.8 n/m. the spring is initially unstr

etched. a constant 23.8 n horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s 2 . 23.8 n 1.35 kg 19.8 n/m find the speed of the block after it has moved 0.236 m from equilibrium if the surface between block and tabletop is frictionless. answer in units of m/s.
Physics
1 answer:
Ket [755]3 years ago
7 0
Horizontal force is applied to the object causing the spring to stretch. the acceleration of gravity is 9.8 m/s
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How does the suns energy contribute to the carbon cycle
netineya [11]

Answer:

Plants are a good starting point when looking at the carbon cycle on Earth. Plants have a process called photosynthesis that enables them to take carbon dioxide out of the atmosphere and combine it with water. Using the energy of the Sun, plants make sugars and oxygen molecules.

8 0
3 years ago
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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length
SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1=L_1

Length of wire 2=L_2

Resistivity of copper wire=\rho_1=1.7\times 10^{-5}\Omega-m

Resistivity of aluminum wire=\rho_2=2.82\times 10^{-5}\Omega-m

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance =\frac{\rho l}{A}

When diameter of wires are same then their cross section area are also same .

l=\frac{RA}{\rho}

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity  is small then the length of wire will be large.

\rho_1

Therefore, L_1>L_2

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66

L_1=1.66L_2

7 0
3 years ago
Can someone answer these
LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

F_airbag = mass * acceleration = 75 kg * 60 * 9.81 mass * acceleration = 44145 newtons

The answer is 4.41 * 10^4

Answer C

Five

This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

Fg = the Force of gravity

Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217  Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

Six

The first thing you should do is derive a general formula for this problem.

The force pulling both masses down is M*g where g is the acceleration due to gravity.

The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a =  [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

4 0
3 years ago
In 10s a car accelerates 4m/s^2 to 60m/s. How fast was the car going before it accelerated?
slega [8]

v2 = v1 + at 60 = v1 + 4*10  v1 = 20 m/s

3 0
3 years ago
A box that has a weight of 40 N is hung from a spring. The spring constant of the spring is 400 N/m. How many centimeters will t
Andrej [43]
The mathematical relationship between force and extension for a spring is F = -kx , where F is the restoring force, k is the spring constant and x is the extension. If a box that weighs 40 N is hung from a spring of content 400 N/m, then the extension is equal to x = -F / k = -400 / 40 = -10 cm. The negative sign simply shows the extension and restoring force are in opposite directions. The answer is 10 cm. 
5 0
3 years ago
Read 2 more answers
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