An object is made from a material that is hard, shiny, and can be hammered into thin sheets. Which of these is another likely property of the material?

Answers

Answer 1

Answer:

Answer:

Good conductor of heat

Explanation:

Because metals are shiny, ductile, malleable, sonorous, good conductors of heat and electricity and have high melting and boiling points

Answer 2

Answer:

Answer: mine is different so im sorry im here for points

Explanation:

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A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3. (a) Draw a force diagram for the balloon. (Submit a file with a maximum size of 1 MB.) (b) Calculate the buoyant force acting on the balloon. (Give your answer to at least three significant figures.) 4159 N (c) Find the net force on the balloon. 1524 N Determine whether the balloon will rise or fall after it is released. The balloon will (d) What maximum additional mass can the balloon support in equilibrium? 155 kg (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? The balloon and its load will remain stationary. The balloon and its load will accelerate downward. The balloon and its load will accelerate upward. (f) What limits the height to which the balloon can rise?

A force of 200N acts on a body that moves along a horizontal plane in the same direction of movement. The body moves 30m. What is the work done by that force?

Answers

Work = Force times Distance

Work = 200 x 30

Work = 6000

The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.

What do you think most likely cause the differences seen in the two graphs for cold spring and normal wearher

Answers

Temperature variations in a graph occur as a result of changing environmental conditions and changing temperature.

what is temperature ?

Temperature is a physical quantity which measures hotness and coldness of a body. Temperature measures the degree of vibration of molecule in a body. Temperature is measured in centigrade (°C), Fahrenheit (°F) and Kelvin (K) in which Kelvin (K) is a SI unit of temperature. Absolute scale of temperature means Kelvin scale of temperature. relation between Kelvin(K) and centigrade (°C), °C= K - 273.15 from equation, 273.15 K means 0 °C, which is freezing point of water (ice). when we give temperature to the body, its molecule or atom absorbs thermal energy and vibrate about their mean position. Amplitude of vibration get increases as we go on increasing temperature and for higher temperature force of attraction between molecules gets weaker. Hence for higher temperature, due to weaken the force of attraction between molecule, solid goes into liquid state. and further increase in temperature liquid goes into gaseous state.

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Answer:

The tempature changes, and the envronment chnages because of this, therefore making tempature changes in a graph.

Explanation:

sorry if this isnt good

An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. Xf-Xi=Vit-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(__s)+1/2a(___s)

Answers

Answer:

Acceleration = 8.27 cm/s²

Explanation:

We are given;

initial velocity; v_i = 10.5 cm/s

Initial position; x_i = 2.72 cm

Time; t = 2.30 s

final position; x_f = 5.00 cm

To find the acceleration, we will make use of the formula;

x_f - x_i = (v_i * t) - (½at²)

Plugging in the relevant values, we have;

5 - 2.72 = (10.5 × 2.3) - (½ × a × 2.3²)

2.28 = 24.15 - 2.645a

24.15 - 2.28 = 2.645a

2.645a = 21.87

a = 21.87/2.645

a = 8.27 cm/s²

Using the kinematic equation, the acceleration of the object was calculated to be approximately8.27 cm/s² given its initial velocity, position, time, and final position.

We are given:

Initial velocity (vᵢ) = 10.5 cm/s

Initial position (xᵢ) = 2.72 cm

Time (t) = 2.30 seconds

Final position ( $x_f$ ) = 5.00 cm

We want to find the acceleration (a) of the object using the kinematic equation:

x₋ᵢ - xᵢ = (vᵢ * t) - (1/2) * a * t²

Now, let's substitute the given values:

5.00 cm - 2.72 cm = (10.5 cm/s * 2.30 s) - (1/2) * a * (2.30 s)²

Simplify the equation:

2.28 cm = 24.15 cm - (1/2) * a * 5.29 s²

Now, isolate 'a' by rearranging the equation:

-1.09 cm = (-1/2) * a * 5.29 s²

To remove the negative sign, multiply both sides by -1:

1.09 cm = (1/2) * a * 5.29 s²

Next, solve for 'a' by multiplying both sides by (2 / 5.29):

a ≈ (1.09 cm) / (2 / 5.29) s²

a ≈ 8.27 cm/s²

So, the acceleration of the object is approximately 8.27 cm/s².

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A car with tires of radius 0.25 m come to a stop from 28.78 m/s (100 km/hr) in 50.0 m without any slipping of tires. Find: (a) the angular acceleration of the wheels; (b) number of revolutions made while coming to rest.

Answers

Answer:

The answer is below

Explanation:

a) Using the formula:

$\omega^2=\omega_o^2+2\alpha \theta\n\n\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\n\theta=angular\ distance\n\nGiven\ that:\n\ninitial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\nfinal/ velocity(v)=0(stop)\n\n\omega=v/r=(28.78m/s)/(0.25m) =115.12\ rad/s,\omega_o=0,\theta=s/r=(50\ m)/(0.25\ m)=200\ rad\n \n\omega^2=\omega_o^2+2\alpha \theta\n\n115.12^2=0^2+2\alpha(200)\n\n2\alpha(200)=13252.6144\n\n\alpha=33.13\ rad/s^2$

$\theta=200\ rad=200\ rad*(1\ rev)/(2\pi\ rad)=31.83\ rev$

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

Answers

Answer:

$y_(max) = 0.829\,m$

Explanation:

Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

$v = \sqrt{(0.8\,(m)/(s))^(2) + 2\cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m)}$

$v = 2.913\,(m)/(s)$

The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

$(3\,kg)\cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m) + (1)/(2)\cdot (3\,kg)\cdot (2.913\,(m)/(s) )^(2) = (3\,kg) \cdot (9.807\,(m)/(s^(2)))\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s) ^(2)$

After some algebraic handling, a second-order polynomial is formed:

$12.728\,J = (1)/(2)\cdot (2000\,(N)/(m) )\cdot (\Delta s)^(2) - (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \Delta s$

$1000\cdot (\Delta s)^(2)-29.421\cdot \Delta s - 12.728 = 0$

The roots of the polynomial are, respectively:

$\Delta s_(1) \approx 0.128\,m$

$\Delta s_(2) \approx -0.099\,m$

The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

$\Delta s \approx 0.128\,m$

The maximum height that the block reaches after rebound is:

$(3\,kg) \cdot (9.807\,(m)/(s^(2)) )\cdot (0.4\,m-\Delta s) + (1)/(2)\cdot (2000\,(N)/(m))\cdot (\Delta s)^(2) = (3\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot y_(max)$

$y_(max) = 0.829\,m$

Answer:

0.81 m

Explanation:

In all moment, the total energy is constant:

Energy of sistem = kinetics energy + potencial energy = CONSTANT

So, it doesn't matter what happens when the block hit the spring, what matters are the (1) and (2) states:

(1): metal block to 0.8 m above the floor

(2): metal block above the floor, with zero velocity ( how high, is the X)

Then:

$E_(kb1) + E_(gb1) = E_(kb2) + E_(gs2)$

$E_(kb1) + E_(gb1) = 0 + E_(gs2)$

$(1)/(2)*m*V_(b1) ^(2) + m*g*H_(b1) = m*g*H_(b2)$

$H_(b2) = (V_(b1) ^(2) )/(2g) + H_(b1)$

Replacing data:

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.8$

$H_(b2) = (0.44^(2) )/(2*9.81) + 0.4$

HB2 ≈ 0.81 m

A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the ______ direction.

Answers

Answer:

Negative z-direction.

Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v* B)$ (or, alternatively, with the left hand rule) that the magnetic force points in the positive z-direction (also taking into account that the charge is positive), so the electric field should be in the negative z-direction to balance it.

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A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.(b) Calculate the magnification.

Somebody tell me what the answer is please