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andrew-mc [135]

3 years ago

8

A force of 6.0 Newtons is applied to a block at rest on a horizontal frictionless surface over a 7.0 meter span. How much energy

is gained by the block?
a) 3.0 m/s

b) 7.0 m

c) 42 J

d) 6.0 N

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

C. 42J

Explanation:

Given data

Force= 6N

Distance= 7m

Required

The energy gained

Work= Force* Distance

Work= 6*7

Work= 42Joules

Hence the option c 42J is correct

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Whats the Independent and Dependent Variables

astra-53 [7]

The independent variable is the type of fuel used and the dependent variable is the speed of the race car. The independent variable could be changed through the experimental process to see its relation with the dependent variable<span>. The dependent variable is the result of the independent variable changes.</span>

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3 years ago

An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field.

aalyn [17]

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

5 0

2 years ago

In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude

melamori03 [73]

Answers:
(a) $B_o = 0.3466$ μT
(b) $\lambda = 0.4488$ μm
(c) f = $6.68 * 10^{14}Hz$

Explanation:
Given electric field(in y direction) equation:
$E_y = 104sin(1.40 * 10^7 x -\omega t)$

(a) The amplitude of electric field is $E_o = 104$ . Hence

The amplitude of magnetic field oscillations is $B_o = \frac{E_o}{c}$
Where c = speed of light

Therefore,
$B_o = \frac{104}{3*10^8} = 0.3466$ μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
$\frac{2 \pi }{\lambda} = 1.40 * 10^7$
$\lambda = \frac{2 \pi}{1.40} * 10^{-7}$
$\lambda = 0.4488$ μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = $6.68 * 10^{14}Hz$

6 0

3 years ago

Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

Simora [160]

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

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Answer:

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