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3 years ago

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HELP ME PLS >>>>>>>>

1 answer:

Margarita [4]3 years ago

7 0

D) a car speeding up may i have brainliest hope this help

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Adding energy or increasing the speed of the particles at very low pressure in a solid would cause

weqwewe [10]

D. sublimation is correct

7 0

2 years ago

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

2 years ago

1. The 10-min unithydrograph for a 2.25 km2 urban catchment is given by:(a) estimate the run off hydrograph for a 10-min rainfal

vivado [14]

Answer:

A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .

Explanation:

Brainlyist

5 0

2 years ago

A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app

zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

Explanation:

We can use the following equation:

$\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta$ (1)

The angular acceleration is:

$a_{tan}=\alpha R$

$\alpha=\frac{1.9}{0.325}$

$\alpha=5.85\: rad/s^{2}$

and the initial angular velocity is:

$\omega_{i}=\frac{v}{R}$

$\omega_{i}=\frac{27.2}{0.325}$

$\omega_{i}=83.69\: rad/s$

Now, using equation (1) we can find the revolutions of the tire.

$0=83.69^{2}-2*25.85 \Delta \theta$

$\Delta \theta=135.47\: rad$

Therefore, the revolutions that each tire makes is:

$\Delta \theta=22\: rev$

I hope it helps you!

6 0

2 years ago

A circular loop of wire with 10 turns, radius 0.241 m and resistance 0.235 Ohms is connected to a 13.1 V power supply. The magne

SVETLANKA909090 [29]

Given Information:

Resistance of circular loop = R = 0.235 Ω

Radius of circular loop = r = 0.241 m

Number of turns = n = 10

Voltage = V = 13.1 V

Required Information:

Magnetic field = B = ?

Answer:

Magnetic field = 0.00145 T

Explanation:

In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B

B = μ₀nI/2r

Where μ₀= 4πx10⁻⁷ is the permeability of free space and current in the loop is given by

I = V/R

I = 13.1/0.235

I = 55.74 A

B = 4πx10⁻⁷*10*55.74/2*0.241

B = 0.00145 T

Therefore, the magnetic field at the center of this circular loop is 0.00145 T

8 0

2 years ago