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vovangra [49]
2 years ago
7

The slope of the line on a speed-time graph tells the speed , true or false ?

Physics
2 answers:
KengaRu [80]2 years ago
8 0
True I hope this helps you out 
shusha [124]2 years ago
4 0
This is true, the slope of the line of a speed graph depending on how high or low would make up for the speed of the line or the object the line is representing.
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How many ounces are in one ton? 16 ounces 125 ounces 2,000 ounces 32,000 ounces
ale4655 [162]

32,000 ounces. hope it helps

5 0
3 years ago
Read 2 more answers
To measure the height of a building without a ruler or tape measure, an engineer drops a rock off the top of the building and fi
serg [7]

The relevant equation we can use in this problem is:

h = v0 t + 0.5 g t^2

where h is height, v0 is initial velocity, t is time, g is gravity

 

Since it was stated that the rock was drop, so it was free fall and v0 = 0, therefore:

h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2

<span>h = 117.77 m</span>

6 0
3 years ago
Am arrow of mass 1000kg is shot into a wooden block of mass 5000lg lying at rest in a smooth surface.If the arrow travels 15m/s
Ber [7]

Answer:

Vf=3

Explanation:

you must first write your data

data before impact

M1=1000 M2=5000

V1=0 m/s V2 =0m/s

data after impact

M1=1000 M2=5000

V1=15m/s V2=?

M1V1 +M2V2=M1V1 +M2V2f

(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf

0=15000+5000Vf

- 15000÷5000=5000Vf÷5000

Vf= -3

Vf =3

6 0
2 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
2 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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