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2 years ago

Which table correctly identifies the abbreviation for SI units of length mass volume and temperature

Physics

1 answer:

SIZIF [17.4K]2 years ago

8 0

Answer: b

Explanation:

Temperature: kelvin

Mass: kilogram

Length: Meter

Volume: Cubic meter

(Apex learning test)

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If the newton is the product of kilograms and meters/second2 what units comprise the pound?

Kobotan [32]

Answer:

Pound is the product of slug and foot/square second.

Explanation:

We are given that

Force=1 N

1N= $1kg\times ms^{-2}$

We have to find the units comprise the pound.

Force=1 Pound

Mass=Slug

Acceleration= $ft/s^2$

Therefore,

1 pound= $1 slug\times fts^{-2}$

Therefore, we can write as 1 pound is equal to the product of slug and ft/square second.

Hence, pound is the product of slug and foot/square second.

6 0

2 years ago

before colliding, the momentum of block A is +15.0 kg m/s. after, block A has a momentum -12.0 kg*m/s. what is the momentum of b

Helen [10]

Answer:

The momentum of block B = 27 Kg m/s

Explanation:

Given,

The initial momentum of block A, MU = 15 Kg m/s

The final momentum of block A, MV = -12 Kg m/s

Consider the block B is initially at rest.

Therefore, the initial momentum of block B, mu = 0

According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.

15 + (0) = (-12) + mv

mv = 15 + 12

= 27 Kg m/s

Hence, the momentum of the block B after impact is, mv = 27 Kg m/s

3 0

2 years ago

A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the

krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒ Ycm = (ms*D + mp*0) / (ms + mp)

⇒ Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0

2 years ago

A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction

aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

W Sinθ = f

mg Sinθ = f

m (9.8) Sin35 = 560

m = 99.63 kg

5 0

2 years ago

A student releases a block of mass m from rest at the top of a slide of height h1. The block moves down the slide and off the en

Nina [5.8K]

Answer:

B) d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance

2) If the height is very small downwards, it touches the earth a little and the horizon is small,

B) to find an equation for horizontal distance (d)

We must maximize the speed at the bottom of the slide let's use energy

Starting point Higher

Em₀ = U = m g h₁

Final point. Lower (slide bottom)

Emf = K + U = ½ m v² + m gh₂

As there is no friction the energy is conserved

mgh₁ = ½ m v² + mgh₂

v² = 2 g (h₁-h₂)

This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)

The distance traveled when leaving the table can be searched with kinematics, projectile launch

x = v₀ₓ t

y = t - ½ g t²

The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero

t = √ 2h₂ / g

We substitute in the other equation

d = √ (2g (h₁-h₂)) √ 2h₂ / g

d = √ (4 h₂ (h₁-h₂))

H = h₁ + h₂

h₁ = H -h₂

d = √ ( 4 h₂ ( H - 2h₂))

Explanation:

7 0

2 years ago