Please answer this im not sure

What is the Kinetic Energy of the glass when it is

poizon [28]

Answer: 197

Explanation:

Because mechanical energy and mass it speeds up

7 0

3 years ago

What is the acceleration of a 10kg pushed by a 5n force

Semenov [28]

Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a

8 0

3 years ago

Suppose that a steel bridge, 1000 m long, were built without any expansion joints. Suppose that only one end of the bridge was h

Stels [109]

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion $\alpha= 10.5\times10^{-6}\ K^{-1}$

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

$\Delta L=L\alpha\Delta T$

Where, $\Delta T$ = temperature difference

$\alpha$ =Coefficient of thermal expansion

L= length

Put the value into the formula

$\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})$

$\Delta L=0.42\ m$

Hence, The difference in the length of the bridge is 0.42 m.

5 0

3 years ago

A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0

3 years ago

A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl

soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

$\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s$

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ = Iω₁ - m v r ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂= I₂ ω₂

I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁ - m v r = I₂ ω₂

Iω₁ - m v r = ( I + m r²) ω₂

Now by putting the all values

Iω₁ - m v r = ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 = ( 0.13 + 0.173 x 0.19²) ω₂

0.325 - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

$\omega_2=\dfrac{2\pi \times N_2}{60}$

N₂=20.05 rpm

3 0

3 years ago