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3 years ago

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Physics

1 answer:

Archy [21]3 years ago

5 0

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is 98 Newtons.

Greetings.

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time

andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

In this case:

x0= 0
v0= 0 because the object is initially stationary
a= 6 $\frac{m}{s^{2} }$
t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 $\frac{m}{s^{2} }$ *(15s)²

Solving:

x=½*6 $\frac{m}{s^{2} }$ *(15s)²

x=½*6 $\frac{m}{s^{2} }$ *225 s²

x= 675 m

The object will travel 675 m during that time.

5 0

3 years ago

What is the mechanical energy of a 500kg rollercoaster car moving with a speed of 3m/s at the top hill that is 30m high

koban [17]

K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J

m*g*h = 500(9.8)(30) = 147000 J

2250 + 147000 = 149250

4 0

3 years ago

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

3 years ago

A single-phase, 125-volt receptacle installed to serve a washing machine in the laundry room of a dwelling unit must be installe

Allisa [31]

A single-phase, 125-volt receptacle installed to serve a washing machine in the laundry room of a dwelling unit must be installed within at least 6 Feet of the intended location of the appliance.

In every kitchen, dining room, library, bedroom or area of dwelling units etc the receptacle units outlet shall be installed with provision specified in article 210.52(A)(1) and through (A)(4) while learning the article attached

Spacing: Receptacles should be installed at the point that wall space is more than 6ft or 1.8m from the receptacle.