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2 months ago

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A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

e lands on a hill 7.5m away. The height at which the projectile lands is most nearly
A- 1.78m
B- 3.10m
C- 5.34m
D- 6.68m
E- 12.0m

2 answers:

ValentinkaMS [3.4K]2 months ago

7 0

Answer:gggffdss

Explanation:cc bbhfewaz

Softa [3K]2 months ago

6 0

$vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7$ Answer:

Explanation:

The range is specified as

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A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.

Keith_Richards [3271]

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

$m = 100 * 11.34 = 1134 g$

$m = 1.134 kg$

Therefore, its weight in air is noted as

$W = mg = 1.134* 9.8 = 11.11 N$

Next, the buoyant force acting on the lead is defined as

$F_B = W - F_{net}$

$F_B = 11.11 - 11 = 0.11 N$

We know that

$F_B = \rho V g$

$0.11 = 1000* V * 9.8$

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

$F_B = W$

$\rho V g = W$

$13* 10^3 * V * 9.8 = 11.11$

$V = 87.2 cm^3$

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

3 0

1 month ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

$\texttt{ }$

Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

$\texttt{ }$

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

1 month ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Softa [3030]

B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.

4 0

1 month ago

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Maru [3345]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The worker is positioned at 2.5 m

7 0

2 months ago

A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At

kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + $\frac{1}{2}$ a $t^{2}$

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + $\frac{1}{2}$ (2.25 x $15.4^{2}$ )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s² (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

$v^{2}$ = $u^{2}$ + 2gs

$0^{2}$ = $34.65^{2}$ + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m + 61.19 m = 328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying $v^{2}$ = $u^{2}$ + 2gs

$v^{2}$ = $0^{2}$ + 2(9.81 x 328)

$v^{2}$ = 0 + 6435.36

v = $\sqrt{6435.36}$ = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0

2 months ago