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2 years ago

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A football player kicks a ball with an initial velocity of 25 m/s at an angle of 53 degrees above the horizontal. The vertical c

omponent of the initial velocity of the ball

1 answer:

Verdich [7]2 years ago

7 0

Answer:

20 m/s

Explanation:

initial velocity, u= 25m/s

u has two components. one in horizontal and in one vertical.

so,

u sin theta = vertical component.

=> u sin theta = 25× 4/5

= 20 m/s

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A 2000 kg car experiences a constant braking force

Marat540 [252]

In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

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then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
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brakes overheat and erupt in flame. Even if the engine is still delivering
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They do not absorb and dissipate any energy when the car is motionless.

4 0

2 years ago

Calculate the hydrostatic difference in blood pressure between the brain and the foot in a person of height 1.93 m. The density

Slav-nsk [51]

Answer:

Explanation:

Given: Density of blood = 1.03 × 10³ Kg/m³, Height = 1.93 m g = 9.8 m/s²

pressure at the brain is equal to atmospheric pressure. = Hydro-static

pressure(ρ₀)

∴ pressure of the foot = pressure of the brain(ρ₀) + ( density of blood × acceleration due to gravity × height)(ρgh)

Hydro-static pressure = pressure at the feet- pressure at the brain(ρ₀)

Hydro-static pressure (Δp) = (ρgh + ρ₀) - ρ₀ = ρgh

Hydro-static pressure = 1.03 × 10³ × 9.8 × 1.93 = 1.948 × 10⁴ Pa

∴ Hydro-static pressure ≈ 1.95 × 10⁴ Pa

3 0

2 years ago

Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn

mafiozo [28]

Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph

Resultant velocity,

$v = \sqrt{v_{2} ^{2}+ v_{1} ^{2} } =\sqrt{63^{2} +53^{2} }$

v = 82.32 mph

5 0

3 years ago

A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto

sesenic [268]

<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float= $l\times b\times h=5.2\times 2.4\times h=12.48h$

Density of water= $10^3kg/m^3$

Weight of water displaced by barge=Buoyant force=-Weight of horse

$Volume\;of\;water\times density\;of\;water\times g=410\times g$

$12.48h\times 1000=410$

$h=\frac{410}{12.48\times 1000}=0.03 m$

1 m=100 cm

$0.03 m=0.03\times 100=3$ cm

Hence, the depth of barge float=3 cm

<h2 />

4 0

3 years ago

Read 2 more answers

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

Read 2 more answers