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3 years ago

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A bullet is fired at an angle θ above the horizontal with an initial velocity of 800 m/s from the top of an 80 m high tower. Wha

t value of θ will give the maximum horizontal range?

1 answer:

Irina-Kira [14]3 years ago

6 0

Answer:

pahingi po ng pic pls para masagutang kopo iyan

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An autographed baseball rolls off of a 0.76 m high desk and strikes the floor 0.61 m away from the desk. how fast was it rolling

zmey [24]

d = distance = 0.76 m <span>
<span>a = acceleration due to gravity = 9.81 m/s^2</span>
u = initial velocity = 0 (as the ball rolls off the table the vertical velocity = 0
t = time = missing so we need to solve it

So we use the equation d = ut + 1/2 at², and ever since u is zero, ut is zero and the equation becomes to d = 1/2 at² and this reorders to t = sqrt (2d/a) = 0.39 seconds.

Since there are no forces performing in the horizontal direction, this means that there is no acceleration in the horizontal direction and consequently the horizontal velocity is persistent. </span>

Velocity = distance/ time.

Horizontal velocity is therefore horizontal distance/time = 0.61 m/0.39s = 1.56 m/s.

<span> </span>

8 0

4 years ago

An engine does 25 J of work and exhausts 20 J of waste heat during each cycle. If the cold-reservoir temperature is 30 ∘C, what

hodyreva [135]

Answer:

T₂=659.25 K

Explanation:

Given that

W= 25 J

Qr = 20 J

T₁ = 20⁰ = 20 +273 = 293 K

The minimum temperature of the hot reservoir = T₂

If the engine is Carnot engine then

Qa= W+ Qr

Qa=25 + 20 J

Qa= 45 J

$\dfrac{Qa}{Qr}=\dfrac{T_2}{T_1}$

$T_2=\dfrac{Qa}{Qr}\times T_1$

$T_2=\dfrac{45}{20}\times 293\ K$

T₂=659.25 K

Therefore the temperature of hot reservoir will be 659.25 K

4 0

4 years ago

47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

$N_1* x = N_2* (L - x)$

also we know that stress at both ends are same

$\frac{N_1}{12} = \frac{N_2}{8}$

$2*N_1 = 3*N_2$

Now from two equations we have

$\frac{3}{2}N_2*x = N_2* (L - x)$

solving above equation we have

$x = \frac{2}{5}L$

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

8 0

3 years ago

A power transmission line is hung from metal towers with glass insulators having a resistance of 1.00×109 Ω . What current flows

Hunter-Best [27]

Answer:

The current flows through the insulator is 2 mA.

Explanation:

Given that,

Resistance $R=1.00\times10^{9}\ \Omega$

Voltage = 200 kV

We need to calculate the current

Using ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Where, I = current

V = voltage

R = resistance

Put the value into the formula

$I=\dfrac{200\times10^{3}}{1.00\times10^{9}}$

$I=0.0002\ A$

$I=2\ mA$

Hence, The current flows through the insulator is 2 mA.

5 0

4 years ago

2. A 200.0-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and th

Neporo4naja [7]

Answer: 1960 N

Explanation:

The bear is sliding down at constant velocity: this means that its acceleration is zero, so the net force is also zero, according to Newton's second law:

$F_{net}=ma=0$

There are two forces acting on the bear: its weight W, pulling downward, and the frictional force Ff, pulling upward. Therefore, the net force is given by the difference between the two forces:

$F_{net}=W-F_f=0$

From the previous equation, we find that the frictional force is equal to the weight of the bear:

$F_f=W=mg=(200.0 kg)(9.8 m/s^2)=1960 N$

8 0

3 years ago