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2 years ago

What is the acceleration of a car that goes from 0 to 20m/s in 7 seconds?

Physics

1 answer:

nadezda [96]2 years ago

6 0

Answer:

2m/s^2

Explanation:

Clculate the acceleration:

V = u +at

20m/s = 0 + a*10s

a = 20m//10s

a = 2m/s²

From the data given , it is not possible to calculate the displacement , because no direction of motion is given

But it is possible to calculate the distance travelled

Distance = ut + ½ *a*t²

distance = 0 + ½ * 2m/s * 10²s

distance = 100m

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what voltage would produce a current of 100A through an aluminum wire that has a resistance of 3.44x10^-4 ohm?

Serga [27]

Voltage = Current x Resistance
Voltage(?) = 100 x 1.98x10^-4 ohms 

Voltage = Current x Resistance 
Voltage(?) = 250 x 2.09x10^-4 ohms 

Voltage = Current x Resistance 
Voltage(?) = 100 x 3.44x10^-4 ohms

7 0

2 years ago

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

A temperature inversion happens when a _____ air layer traps pollution close to the surface of the earth.

LenaWriter [7]

When warm air layer traps pollution that is within the earth's surface, a temperature inversion usually takes place. In addition, at temperature inversion happens when the cold air overlays the warm air in the troposphere, which does not usually happen since warm air rises and cold air sinks by principle.

8 0

3 years ago

What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other

Alex73 [517]

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

4 0

2 years ago

The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie

Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I' and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0

2 years ago

Read 2 more answers