Answer:
Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.
The volume of a solid object tends to increase as its temperature increases. Which quantities determine how large the change in volume is
Answer:
I. The change in temperature of the object.
II. The initial volume of the object.
Explanation:
Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.
Hence, the quantities which determine how large the change in volume of a solid object is includes;
I. The change in temperature of the object.
II. The initial volume of the object.
This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.
In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;
Solid. Liquid. Gas.which of the following are not units used to measure energy?
a. joules
b. newtons
c. BTU
d. calories
Answer:
The BTU, or British thermal unit, is actually a measure of heat.
A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.
Answer:
-22.1
Explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.
A hockey puck slides across the ice and eventually comes to a stop. Why did the puck stop?
Answer: the total energy of the puck, ice surface, and surrounding air decreases to zero
Explanation:
A hockey puck slides across the ice and eventually comes to a stop because of friction between surface of the puck and ice surface.
What is Friction ?Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force. frictional force is directly proportional to the Normal(N). i.e. [tex]F_{fri}[/tex] ∝ N
[tex]F_{fri}[/tex] = μN where μ is called as coefficient of the friction. It is a dimensionless quantity.
When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.
Frictional force is equal to
[tex]F_{fri}[/tex] = μmg
When hockey puck slides across the ice, friction between surface of puck and surface of ice produces resistance to the motion of the puck, due to resistance puck slow down slowly and eventually come to a stop. Motion and frictional force are opposite to each other. we can calculated the exact value of frictional force when we know the coefficient of friction between puck and ice surface.
hence due to friction, puck come to a stop
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A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?
(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)
The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.
The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.
From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.
where;
|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile∴
Using trigonometric approach for ΔOAP;
[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]
[tex]\mathbf{AP =OP\times sin \theta}}[/tex]
AP = 200 × sin 30°
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When catching a baseball, a catcher's glove moves by 11 cm along the line of motion of the ball. If the baseball exerts a force of 464 N on the glove, how much work is done by the ball
Answer:
51.04 joules
Explanation:
The movement by the gloves = 11cm
Force by the baseball = 464N
First of all we are going to convert cm to metres
11cm = 11/100 meters = 0.11m
The formula for workdone is given as:
W = f x d
W = workdone
F = force
D = distance
Workdone = 464 x 0.11
= 51.04 joules
The workdone by the ball is 51.04 joules
Thank you
A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel the fastest? Aluminum Copper Lead Brass (70% Cu, 30% Zn) 6,420 4,700 5,010 1,960 Speed of sound (m/s) O
A. Brass O
B. Aluminum
C. Copper O
D. Lead
Answer:
Aluminum
Explanation:
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Answer:aluminum
Explanation:
4. Friction is required for :
(a) Fast movement
(0) Both of the above
(b) Stopping the objec
(d) None of the above
Answer:
b) stopping the object
Explanation:
Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding