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N76 [4]
4 years ago
10

Can a vector have nonzero magnitude if a component is zero?

Physics
1 answer:
Ksivusya [100]4 years ago
4 0
The vector may have at least two components. These are the x-component and the y-component. The resultant vector (or vector, as referred in this item) is the summation of the components. With this, it is still possible for the vector or resultant to have a nonzero magnitude even if one of its components is zero. 
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Two blocks of masses 8 kg and 4.3 kg are placed on a horizontal, frictionless surface. A light spring is attached to one of them
oee [108]

Answer:

The velocity of other mass is 3.60 m/s.

Explanation:

Given that,

Mass of first block = 8 kg

mass of second block = 4.3 kg

Speed = 6.7 m/s

We need to calculate the speed of first mass

Using conservation of momentum

(m_{1}+m_{2})u=m_{1}v_{1}+m_{2}v_{2}

where, m₁ =mass of first block

m₂ =mass of second block

m₁ =mass of first block

v₂ =speed of second block

Put the value into the formula

8+4.3\times0=8\times v_{1}+4.3\times6.7

v_{1}=\dfrac{-4.3\times6.7}{8}

v_{1}=-3.60\ m/s

Negative sign represent the opposite direction of initial value.

Hence, The velocity of other mass is 3.60 m/s.

3 0
3 years ago
How strong an electric field is needed to accelerate electrons in an X-ray tube from rest to one-tenth the speed of light in a d
Bond [772]

Answer:

E= 50.1*10³ N/C

Explanation:

Assuming no other forces acting on the electron, if the acceleration is constant, we can use the following kinematic equation in order to find the magnitude of the acceleration:

vf^{2} -vo^{2}  = 2*a*x

We know that v₀ = 0 (it starts from rest),  that vf = 0.1*c, and that x = 0.051 m, so we can solve for a, as follows:

a = \frac{vf^{2}}{2*x} = \frac{(3e7 m/s)^{2} }{2*0.051m} =8.8e15 m/s2

According to Newton's 2nd Law, this acceleration must be produced by a net force, acting on the electron.

Assuming no other forces present, this force must be due to the electric field, and by definition of electric field, is as follows:

F = q*E (1)

In this case, q=e= 1.6*10⁻19 C

But this force, can be expressed in this way, according Newton's 2nd Law:

F = m*a (2) ,

where m= me = 9.1*10⁻³¹ kg, and a = 8.8*10¹⁵ m/s², as we have just found out.

From (1) and (2), we can solve for E, as  follows:

E=\frac{me*a}{e} =\frac{(9.1e-31 kg)*(8.8e15m/s2)}{1.6e-19C} = 50.1e3 N/C

⇒ E = 50.1*10³ N/C

3 0
4 years ago
Which is a warning sign that someone needs the help of a mental health professional
REY [17]

Answer:

Suicidal Behavior

Explanation:

If someone is being suicidal, they are in need of mental support.

4 0
3 years ago
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I need help solving this
Alja [10]

Answer: let me think

Explanation:

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
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