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ikadub [295]
3 years ago
11

A sports car accelerates at 2.80 m/s2 along a straight road. At t1 = 5.00s and t2 = 5.50 s it passes two marks that are 35.0 m a

part. What is the car's velocity at t0 = 0 s?​
Physics
1 answer:
harina [27]3 years ago
7 0

Explanation:

Using kinematics,

s = ut + 0.5at².

We have s = 35.0m, t = 0.50s, a = 2.80m/s².

Substitute in the variables, we have u = 69.3m/s.

This u is only the initial velocity at t1. To find the velocity at t0, we apply a different kinematics equation with different values:

We have t = 5.00s, a = 2.80m/s², v = 69.3m/s.

Using kinematics,

v = u + at.

Substitute in the variables, we have u = 55.3m/s.

Hence the car's initial velocity is 55.3m/s.

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The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

3 0
3 years ago
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Sal sprinted 40 m to the right in 5.5s what is his average velocity
PolarNik [594]

Answer: 7.27 m/s

Explanation:

6 0
3 years ago
A boy and his younger sister are at the zoo on a hot day. They each buy a cold lemonade. The boy buys a large lemonade, and his
FromTheMoon [43]

Answer: A and C (i took the test)

Explanation: Hope this helps:)

7 0
4 years ago
The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.
klio [65]
The relationship between frequency and wavelength for an electromagnetic wave is
c=f \lambda
where
f is the frequency
\lambda is the wavelength
c=3 \cdot 10^8 m/s is the speed of light.

For the light in our problem, the frequency is f=1.20 \cdot 10^{13} s^{-1}, so its wavelength is (re-arranging the previous formula)
\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}=  2.5 \cdot 10^{-5}m
8 0
3 years ago
The space shuttle orbits 310 km above the surface of the Earth.
MAVERICK [17]

Answer:

44.7 N

Explanation:

The gravitational force between the objects is given by:

F=G\frac{mM}{r^2}

where

G is the gravitational constant

m and M are the masses of the two objects

r is the distance between the centres of the two objects

In this problem, we have:

m=5.0 kg is the mass of the sphere

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6370 km is the Earth's radius, while h=310 km is the altitude of the sphere, so the distance of the sphere from Earth's centre is

r=6370 km+310 km=6680 km=6.68\cdot 10^6 m

Substituting into the equation, we find

F=(6.67\cdot 10^{-11})\frac{(5.0 kg)(5.98\cdot 10^{24} kg)}{(6.68\cdot 10^6 m)^2}=44.7 N

8 0
3 years ago
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