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3 years ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

Physics

1 answer:

sineoko [7]3 years ago

5 0

Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

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Answer:

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Explanation:

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gamma rays

X-rays

ultraviolet radiation

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It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f

goldenfox [79]

Answer:

B) the change in momentum

Explanation:

Impulse is defined as the product between the force exerted on an object (F) and the contact time ( $\Delta t$ )

$I=F \Delta t$

Using Newton's second law (F = ma), we can rewrite the force as product of mass (m) and acceleration (a):

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However, the acceleration is the ratio between the change in velocity ( $\Delta v$ ) and the contact time ( $\Delta t$ ): $a=\frac{\Delta v}{\Delta t}$ , so the previous equation becomes

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3 years ago

A wide sloping deposit of sediment formed where a stream leaves a mountain range is called

tatuchka [14]

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3 years ago

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

$T = 2 \pi \sqrt\frac{m}{k}$

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

$T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s$

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

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3 years ago