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Vanyuwa [196]
4 years ago
10

True or false when a person swims through water no work is done

Physics
2 answers:
ozzi4 years ago
8 0

Answer:

False

Explanation:

First we need to know the definition of work.

Work is done when a body moves through a distance with an applied force.

When swimming in water, you exert a force while moving through a distance. Therefore, work is done!

sweet [91]4 years ago
4 0
False, work is when a force is exerted
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At what height must a 6-kilogram object be placed to have 12 joules of
Serga [27]

Answer:

0.2m

Explanation:

Using E=mgh

12=6*9.8*h

h=0.2m (1 d.p.)

4 0
4 years ago
If you measure the mass of a model car to be 230 grams, but the actual mass is 218 grams, what is your percent error?
konstantin123 [22]

Mass of a model car you measured = 230 grams

Actual mass = 218 grams

Percent error = ?

First subtract the actual value from the value you measured,

230 grams – 218 grams = 12 grams

Now divide 12 grams by the actual mass that is 218 grams:

12 / 218 = 0.055                             

Now multiply 0.055 with 100 to get the percent error;

0.055 x 100 = 5.5%

So, the Percent error is 5.5%.

8 0
4 years ago
A mass of 10.0 kg is in a gravitational field of 3.50 N/kg. What force acts on the mass?
mash [69]

Answer:

Force=35 N

Explanation:

Given data

mass m=10.0 kg

Gravitational field E=3.50 N/kg

To find

Force

Solution

From definition of gravitational field intensity.

E_{gravitational-field }=\frac{Force}{mass}\\  E=F/m\\F=mE\\F=(10.0 kg)*(3.50 N/kg)\\F=35N

6 0
4 years ago
Read 2 more answers
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How m
serious [3.7K]

Answer:920.31 J

Explanation:

Given

Volume of water (V)=14.1 cm^3

mass(m)=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm

Temperature =8.4^{\circ} C

Final Temperature =-7.2 ^{\circ}C

specific heat of water(c)=4.184 J/g-^{\circ}C

Therefore heat required to removed is

Q=mc(\Delta T)

Q=14.1\times 4.184\times (8.4-(-7.2))

Q=920.31 J

3 0
4 years ago
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