All categories

3 years ago

Please Help me in this question..

Physics

1 answer:

Vlad [161]3 years ago

8 0

Answer:

1 different

Explanation:

2 light

3 sunny

4 a shadow

5dark

<h2 />

You might be interested in

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete Ho

zimovet [89]

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal = 30 degree

tension in the string = 4.5 N

WE KNOW THAT

Work = force * distance

horizontal force = $Tcos\theta = 4.5*cos30 = 3.89 N$

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

3 0

3 years ago

What is the speed of a beam of electrons when under the simultaneous influence of E = 1.64×104 V/m B = 4.60×10−3 T Both fields a

andrezito [222]

Answer: v = 3.57×10^6 m/s; R = 4.42×10^-3m; T = 7.78×10^-9 s

Explanation:

Magnetic force(B) = 4.60×10^-3 T

Electric force(E) = 1.64×10^4 V/m

Both forces having equal magnitude ;

Magnetic force = electric force

qvB = qE

vB = E

v = (1.64×10^4) ÷ (4.60×10^-3)

v = 3.57×10^6 m/s

2.) Assume no electric field

qvB = ma

Where a = v^2 ÷ r

R = radius

a = acceleration

v = velocity

qvB = m(v^2 ÷ R)

R = (m×v) ÷ (|q|×B)

q=1.6×10^-19C

m = 9.11×10^-31kg

R = (9.11×10^-31 * 3.57×10^6) ÷ (1.6×10^-19 * 4.6×10^-3)

R = 32.5227×10^-25 ÷ 7.36×10^-22

R = 4.42×10^-3m

3.) period(T)

T = (2*pi*R) ÷ v

T = (2* 4.42×10^-3 * 3.142) ÷ (3.57×10^6)

T = (27.775×10^-3) ÷(3.57×10^6)

T = 7.78×10^-9 s

6 0

4 years ago

Calculate the impulse of a 1kg box that starts from rest and accelerates to 50m/s over a period of 10 seconds.

Tomtit [17]

Answer:

please find attached pdf

Explanation:

Download pdf

5 0

3 years ago

you hold a metal block of mass 40 kg above your head at a height of 2 m how much work is done bu gravity if you let the block fa

zzz [600]

Answer:

784 J

Explanation:

Work is force times distance.

W = Fd

W = mgd

W = (40 kg) (9.8 m/s²) (2 m)

W = 784 J

8 0

4 years ago

Please help.

Alex787 [66]

Explanation:

mathematically,

$kinetic \: energy \: = \frac{1}{2} m {v}^{2}$

where m= mass

and v=velocity

given that,

kinetic energy=493900J

mass =5040kg

making velocity the subject

$v = \sqrt{ \frac{2 \times kinetic \: energy}{mass} }$

substitute their values in the formula

$v = \sqrt{ \frac{2 \times 493900}{5040} }$

$v = \sqrt{ \frac{987800}{5040} }$

$v = \sqrt{196}$

$v = 14m {s}^{ - 1}$

Thus the velocity is 14meters per second

6 0

3 years ago

Please Help me in this question..​

Please Help me in this question..