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Alenkasestr [34]
3 years ago
7

If a ball is launched upward with a vertical velocity of 50 m/s and a horizontal velocity of 10 m/s, what will the vertical disp

lacement be 5 seconds later?
Physics
1 answer:
Delicious77 [7]3 years ago
8 0

Answer: The vertical displacement will be 127.5 meters

Step by step:

Since the question asks for the vertical displacement only, it is sufficient to limit our calculation to the vertical components of the problem.

We choose to measure the displacement on a vertical axis d with an initial coordinate 0 coinciding with the launch of the ball. There is an initial horizontal velocity v_0 of 50 m/s and the gravitational deceleration g acting in the direction opposite to the initial velocity. The formula for displacement of a body subject to an deceleration with an initial velocity (at time t=5 seconds) is:

d(t) = -\frac{1}{2}gt^2+v_0t+d_0=-\frac{1}{2}9.8 \frac{m}{s^2}t^2+50\frac{m}{s}\cdot t + 0m\\d(5) = -\frac{1}{2}9.8 \frac{m}{s^2}5^2 s^2+50\frac{m}{s}\cdot 5s=127.5m

After 5 seconds of the launch, the ball will be at the point 127.5 meters from the origin on the vertical axis (vertical displacement)


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Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
photoshop1234 [79]

Answer:

7.55 g

Explanation:

Given that:

Heat of fusion = 333kj/kg

Heat capacity, c = 4190 j/kg /k

The Number of grams of ice that will melt can be represented as y:

Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change

y * 333 * 10^3 J = (4190) * (6 - 0)

333000y = 25140

y = 25140 / 333000

y = 0.0754954 kg

y = 0.0754954 * 100

y = 7.549 g

Hence, Number of grams of ice that will melt = 7.55 g

4 0
3 years ago
What is an example of density
Taya2010 [7]
Ice floating on water and Ice floats on water because it is less dense than water. Things that are less dense float on top of things that are more dense because molecules in ice are further apart than molecules in water.
5 0
3 years ago
HURRY WILL GIVE BRAINLIEST
andrey2020 [161]

The amount of energy needed to melt the ice is 1670 J

Explanation:

The amount of thermal energy needed to completely melt a certain amount of solid substance (already at melting point) is given by

Q=m\lambda_f

where

m is the amount of the substance

\lambda_f is the specific latent heat of fusion of the substance

Here, the mass of the ice is

m = 5 g = 0.005 kg

And the specific latent heat of fusion of ice is

\lambda_f = 334,000 J/kg

Assuming that the ice is already at melting point, the thermal energy needed to melt the ice is

Q=(0.005)(334,000)=1670 J

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0
3 years ago
What must be true in order for magnetism to induce an electric current in a wire?
worty [1.4K]

Answer:

The wire must be part of a closed circuit.

Explanation:

A closed circuit in electric is what we called a wire that has a complete electrical connection in which electricity flows, where electric current flow from one end to another, in order for magnetism to induce an electric current the circuit has to be closed, or no current will flow.

7 0
2 years ago
Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m
Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

A=\frac{1-V_{1}^{2}  }{E_{1} }-\frac{1-V_{2}^{2}  }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}

a) The maximum contact pressure is:

P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} })  }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa

The width of contact is:

b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} })  } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

T = 0.3*P = 0.3 * 274.58 = 82.37 MPa

At a distance of

0.8*b = 0.8*0.029 = 0.023 mm

6 0
3 years ago
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