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jolli1 [7]
3 years ago
15

A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio

n?
Physics
1 answer:
ira [324]3 years ago
6 0
According to Newton's 2nd law of motion:
F = m * a where F is the force applied in Newtons, m is the mass of the object in kg, and a is the acceleration of the object in m/s^{2}.
Therefore the force applied in this situation is simply:
F = 6 kg * 2.3 m/s^{2} = 13.8 N
Hope this helps!
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1. What is the arrangement of the outer planets? 2. What effect does their placement have the planets?
Rasek [7]
<span>the arrangement of the outer planets is 
</span>1. Mercury 
<span>2. Venus </span>
<span>3. Earth </span>
<span>4. Mars </span>
<span>5. Jupiter </span>
<span>6. Saturn </span>
<span>7. Uranus </span>
8. Neptune
the inner most of the outer plannets is jupitor it is followed by saturn uranus and neptune
4 0
3 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
2 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
Find the velocity, acceleration, and speed of a particle with the given position function. r(t = t2i 6tj 4 ln t k
artcher [175]
1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k

2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2

Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)

For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.
6 0
3 years ago
Please help!!~
lisov135 [29]

Answer:

C. Waves transfer energy, but not matter.

Explanation: hope this helps :)

7 0
2 years ago
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