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4 years ago

A Roman centurion fires off a vat of burning pitch from

Physics

2 answers:

EastWind [94]4 years ago

6 0

Answer:

(a) 1.11sec

(b) 14.37m/s

Explanation:

U = 18m/s, A = 37°, g = 9.8m/s^2

(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec

(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s

garik1379 [7]4 years ago

5 0

Answer:

a 5.995m

b 14.54m/s

C 31.84m

Explanation:

H= u^2 sin^2x/2g

H= 18.2^2 sin^2×37/2×10

H= 331.24× 0.362/20

H= 5.995m

Horizontal component of the Vat

Ux= ucos x ; x is the angle

Ux= 18.2 cos 37

Ux= 14.54m/s

Range = u^2 sin2x/g

18.2^2 sin 74/10

Range = 318.40/10

Range = 31.84m

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Describe why wearing a long-sleeved white T-shirt can help you feel cooler on a hot day than wearing a short-sleeved black T-shi

MAVERICK [17]

Answer:

Explanation:

In a white T-shirt its is not as dark of a color as black so when the sun hits it you won't get hot as much or as quick. If you wear a back shirt you will most likely be hotter because since it is a darker color when the sun hits it it make you hotter and you heat up quicker. You still might feel hot in a white T-shirt because long sleeve shirts trap in your body heat making you hotter because the heat has nowhere else to go.

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3 years ago

A block of mass m=2.20m=2.20 kg slides down a 30.0^{\circ}30.0

Xelga [282]

Answer:

$v_m \approx -4.38\; \rm m \cdot s^{-1}$ (moving toward the incline.)

$v_M \approx 4.02\; \rm m \cdot s^{-1}$ (moving away from the incline.)

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

If $g = 9.81\; \rm m \cdot s^{-2}$ , the potential energy of the block of $m = 2.20\; \rm kg$ would be $m \cdot g\cdot h = 2.20\; \rm kg \times 9.81\; \rm m \cdot s^{-2} \times 3.60\; \rm m \approx 77.695\; \rm J$ when it was at the top of the incline.

If friction is negligible, all these energies would be converted to kinetic energy when this block reaches the bottom of the incline. There shouldn't be any energy loss along the horizontal surface, either. Therefore, the kinetic energy of this $m = 2.20\; \rm kg\!$ block right before the collision would also be approximately $77.695\; \rm J$ .

Calculate the velocity of that $m = 2.20\; \rm kg$ based on its kinetic energy:

$\displaystyle v_m(\text{initial}) = \sqrt{\frac{2\times (\text{Kinetic Energy})}{m}} \approx \sqrt{\frac{2 \times 77.695\; \rm J}{2.20\; \rm kg}} \approx 8.4043\; \rm m \cdot s^{-1}}$ .

A collision is considered as an elastic collision if both momentum and kinetic energy are conserved.

Initial momentum of the two blocks:

$p_m = m \cdot v_m(\text{initial}) \approx 2.20\; \rm kg \times 8.4043\; \rm m \cdot s^{-1} \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

$p_M = M \cdot v_M(\text{initial}) \approx 2.20\; \rm kg \times 0\; \rm m \cdot s^{-1} \approx 0\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right before the collision: approximately $18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Sum of the momentum of each block right after the collision: $(m\cdot v_m + m \cdot v_M)$ .

For momentum to conserve in this collision, $v_m$ and $v_M$ should ensure that $m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1}$ .

Kinetic energy of the two blocks right before the collision: approximately $77.695\; \rm J$ and $0\; \rm J$ . Sum of these two values: approximately $77.695\; \rm J\!$ .

Sum of the energy of each block right after the collision:

$\displaystyle \left(\frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2\right)$ .

Similarly, for kinetic energy to conserve in this collision, $v_m$ and $v_M$ should ensure that $\displaystyle \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J$ .

Combine to obtain two equations about $v_m$ and $v_M$ (given that $m = 2.20\; \rm kg$ whereas $M = 7.00\; \rm kg$ .)

$\left\lbrace\begin{aligned}& m\cdot v_m + m \cdot v_M \approx 18.489\; \rm kg \cdot m \cdot s^{-1} \\ & \frac{1}{2}\, m \cdot {v_m}^2 + \frac{1}{2}\, M \cdot {v_M}^2 \approx 77.695\; \rm J\end{aligned}\right.$ .

Solve for $v_m$ and $v_M$ (ignore the root where $v_M = 0$ .)

$\left\lbrace\begin{aligned}& v_m \approx -4.38\; \rm m\cdot s^{-1} \\ & v_M \approx 4.02\; \rm m \cdot s^{-1}\end{aligned}\right.$ .

The collision flipped the sign of the velocity of the $m = 2.20\; \rm kg$ block. In other words, this block is moving backwards towards the incline after the collision.

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3 years ago

In 8.5 s a fisherman winds 2.4 m of fishing line onto a reel whose radius is 3.0 cm (assumed to be constant as an approximation)

SSSSS [86.1K]

Answer:

9.412 rad/s.

Explanation:

Velocity is the rate of change of an object's position.

V = x/t

Where x is the distance in m

= 2.4 m

t is time taken in s

= 8.5 s

V = 2.4/8.5

= 0.2824 m/s.

Equating linear velocity and angular velocity,

V = ω*r

Where,

ω Is the angular speed in rad/s

r is the radius of the circle in m

= 3 cm

= 3cm * 1m/100 cm = 0.03 m

ω = V/r

= 0.2824/0.03

= 9.412 rad/s.

4 0

4 years ago

When running a 100 meter race Trevor reaches his maximum speed when he is 40 meters from the starting line, and 6 seconds have e

Virty [35]

Answer:

The max speed is 8m/s, 3 seconds at this speed gives 24mts, at 9 seconds has traveled 64mts and at 7,5 52mts

Explanation:

The max speed can be calculated since, at 6 seconds is at 40 mts from the starting point and at constant speed is at 72 mts 10 seconds after the start of the race, then the constant max speed is (72-40)mts /(10-6)seconds=8 m/s.

the distanced travel at this speed is obtained by multipliying by 3 seconds. that is 8m/s*3s=24 mts. the distance at 9 seconds is 40mt of initial 6 plus 24 additional, that is 64mts and at 7,5 is the 40mts initial plus the 13 additional from 8m/s*1,5s = 12mt, that is 52mts.

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