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emmainna [20.7K]
2 years ago
12

A 5.6 g marble is fired vertically upward using a spring gun. The spring must be compressed 6.4 cm if the marble is to just reac

h a target 15 m above the marble's position on the compressed spring. What is the change in the gravitational potential energy of the marble-Earth system during the 15 m ascent
Physics
1 answer:
N76 [4]2 years ago
5 0

Answer: im not gonna give i to you just do 15+15=_+ 5.6+6.4 easy

Explanation: i took the test and got a 100%

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A 50.0 N box sliding on a rough horizontal floor, and the only horizontal force acting on it is friction. You observe that at on
Vikki [24]

Answer:

-4.0 N

Explanation:

Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):

F_f = ma (1)

We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

m=\frac{W}{g}=\frac{50.0 N}{9.8 m/s^2}=5.1 kg

And we can fidn the acceleration by using the formula:

a=\frac{v-u}{t}

where

v = 0 is the final velocity

u = 1.75 m/s is the initial velocity

t = 2.25 s is the time the box needs to stop

Substituting, we find

a=\frac{0-1.75 m/s}{2.25 s}=-0.78 m/s^2

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)

Therefore, substituting into eq.(1) we find the force of friction:

F_f = (5.1 kg)(-0.78 m/s^2)=-4.0 N

Where the negative sign means the direction of the force is opposite to the motion of the box.

6 0
3 years ago
What remains constant during the trajectory of an object
S_A_V [24]

Answer:

projectile

Explanation:

3 0
3 years ago
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
Static electricity is the
guapka [62]
Unlike charges attracting each other
4 0
3 years ago
Read 2 more answers
You need to design a 60.0-Hz ac generator that has a maximum emf of 5500 V. The generator is to contain a 150-turn coil that has
olchik [2.2K]

Answer:

So magnetic field will be equal to 0.1144 T  

Explanation:

We have given frequency f = 60 Hz

Maximum emf e = 5500 volt

Number of turns N = 150

Area A=0.85m^2

Emf generated in ac generator is given e=NBA\omega sin(\omega t)

For maximum emf sin(\omega t)=1

So maximum emf will be equal to e=NBA\omega

B=\frac{e}{NA\omega }=\frac{5500}{150\times 2\times 3.14\times 60\times 0.85}=0.1144T

So magnetic field will be equal to 0.1144 T

3 0
3 years ago
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