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leonid [27]
2 years ago
12

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

. Determine the projectiles horizontal displacement if the total time of flight is 5 seconds
Physics
1 answer:
Artyom0805 [142]2 years ago
5 0

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

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A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
Anna35 [415]

Answer:

a. The temperature of the copper changed more than the temperature of the water.

Explanation:

Because we're only considering the isolated system cube-water, the heat of the system should be constant, that implies the heat the cube loses is equal the heat the water gains (because by zero law of thermodynamics heat (Q) flows from hot body to cold body until reach thermal equilibrium and T1>T2). So:

Q_{cube}=Q_{water} (1)

But Q is related with mass (m), specific heat (c) and changes in temperature (\varDelta T)in the next way:

Q=cm\varDelta T(2)

Using (2) on (1):

c_{cooper}*m_{cooper}*\varDelta T_{cooper}=c_{water}*m_{waterer}*\varDelta T_{water}

(10g)(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(10g)(4.186 \frac{J}{g\,C})(\varDelta T_{water})

(0.385 \frac{J}{g\,C})(\varDelta T_{cooper})=(4.186 \frac{J}{g\,C})(\varDelta T_{water})

Because we have an equality and 0.385 < 4.186 then \varDelta T_{cooper}>\varDelta T_{waterer} to conserve the equality

4 0
2 years ago
The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10–15 s–1 at 298 K and a rate constant of 8
netineya [11]

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})

Solving for Ea as:

Ea = 159982.23 J /mol

1 J/mol = 10⁻³ kJ/mol

Ea = 159.98 kJ/mol

7 0
2 years ago
When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j
garri49 [273]
We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!
8 0
2 years ago
Read 2 more answers
The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

r = 1.08  × 10⁷ m, G = 6.67  × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08  × 10⁷ )²/75( 6.67  × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0
2 years ago
 If the gauge pressure of a gas is 114 kPa, what is the absolute pressure?
Anastasy [175]

Answer:

D. 214 kPa

Explanation:

The absolute pressure is given by:

p = p_a + p_g

where

p is the absolute pressure

p_a \sim 100 kPa is the atmospheric pressure

p_g is the gauge pressure

In this problem, we have

p_g = 114 kPa

So, the atmospheric pressure is

p = 100 kPa + 114 kPa = 214 kPa

4 0
2 years ago
Read 2 more answers
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